Unit 01 Some Basic Concepts of Chemistry
Matter and It’s Nature
Anything which has mass and occupies space is called matter.
Three physical states of matter are solid, liquid and gas.
In solids, molecules are held very close to each other and they have definite shape and definite volume.
In liquids, molecules are close to each other but move around, they have definite volume but not shape.
In gases, molecules are far apart and have neither definite volume nor shape.
At bulk level, matter can be divided in to pure and impure substances.
Elements and compounds are pure substances whereas mixtures fall in the category of impure substances.
Mixtures are further classified as heterogeneous (with non uniform composition) and homogenous mixtures (with uniform composition).
Elements are made up of identical atoms with the same atomic number.
Dalton’s atomic theory
All matter is made of atoms which are indivisible and indestructible particles.
All the atoms of a given element are identical, however, atoms of different elements have different masses and different chemical roperties.
Compounds are formed by the combination of different atoms in the ratio of small whole numbers.
Atoms are neither created nor destroyed in the course of an ordinary chemical reaction.
Atoms and molecules
An atom is the smallest particle of an element which can take part in a chemical reaction. It may or may not be capable of independent existence.
A molecule is the smallest particle of an element or a compound which is capable of independent existence.
Elements such as oxygen, hydrogen, etc. exist in forms of molecules.
When two or more atoms of different elements combine, the molecule of a compound is obtained such as water, ammonia, carbon-dioxide, etc.
The atoms of different elements are present in a compound in a fixed ratio which is characteristic of a particular compound.
The constituents of a compound cannot be separated into simpler substances by physical methods.
Physical quantities and their measurements in chemistry
Mass - quantity of matter present in a substance.
Weight- force exerted by gravity on an object
Volume-the space occupied by a substance, or enclosed within a container
Density - mass per unit volume
Temperature-the degree or intensity of hotness of a substance
Pressure - force exerted per unit area on or against an object
| Physical Quantity | S.I. Units |
|---|---|
| Mass | Kilogram (kg) |
| Weight | Newton (N) |
| Volume | $\mathrm{m}^{3} ;$ Litre (L) |
| Density | $\mathrm{kg} / \mathrm{m}^{3}$ |
| Temperature | Kelvin (K) |
| Pressure | Pascal $\left(\mathrm{Nm}^{-2}\right.$ or Pa) |
Precision and accuracy
Precision refers to the closeness of various measurements for the same quantity.
Accuracy is the agreement of a particular value to the true value of the result.
Significant figures
All non zero digits are significant.
The zeros to the right of the decimal point or zeros between two non-zero digits are significant.
Zeros to the left of the first non-zero digit in a number smaller than one are not significant.
In addition or subtraction, the result should be mentioned in the same number of decimal place as that of the term with the least ecimal places.
In multiplication and division, the result should be mentioned in same number of significant figures as the least precise term used in calculation.
Dimensional analysis
While calculating one needs to convent units from one system to other. The method used to accomplish this is called factor label method or unit factor method or dimensional analysis.
Ex: 1 inch $=2.54 \mathrm{~cm}$
$$ \frac{1 \text { inch }}{2.54 \mathrm{~cm}}=1=\frac{2.54 \mathrm{~cm}}{1 \mathrm{inch}}=1 $$
Both these are called unit factors
Laws of Chemical Combinations
Law of Conservation of Mass :
Mass is neither created nor destroyed in chemical reactions. For example
$$ \begin{aligned} & 2 \mathrm{H} _{2}(\mathrm{~g})+\mathrm{O} _{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H} _{2} \mathrm{O}(\mathrm{g}) \\ & 4 g \qquad\qquad 32 g \quad\quad\qquad 36 g \end{aligned} $$
Mass remains conserved, i.e., mass of reactant $=$ mass of product.
Law of Multiple Proportions :
If two elements combine in different ways to form different compounds, the masses of one element that combine with a fixed mass of the other element are in ratio of small whole numbers. For example
| $\mathrm{NO}$ | $\mathrm{NO} _{2}$ | |
| Mass ratio of N and $\mathrm{O}$ | $14: 16$ | $14: 32$ |
| $7: 8$ | $7: 16$ |
Ratio of masses of 0 that combines with a fixed mass of $\mathrm{N}(7 \mathrm{~g}) 8: 16$ or $1: 2$.
Law of Constant Proportions :
Different samples of a pure chemical substance always contain the same proportions of elements by mass. e.g. every sample of water $\left(\mathrm{H} _{2} \mathrm{O}\right)$ contains 1 part hydrogen and 8 parts oxygen by mass (i.e., 1:8).
Law of Reciprocal Proportions :
When definite mass of an element A combines with two different elements $B$ and $C$ to form two compounds and if $B$ and $C$ also combine to form a compound, their combining masses bear a simple ratio to the masses of $B$ and $C$ which combine with a constant mass of A.
Gay Lussac’s Law :
According to this law, under similar conditions of temperature and pressure, gases react with each other in a simple ratio of their volumes and if the product is also gaseous, it also bears a simple ratio with the volumes of reactants.
Avogadro’s Law :
Equal volume of all gases under the same conditions of temperature and pressure contain equal number of molecules.
Atomic mass
Relative atomic mass of an element is defined as the ratio of the mass of one atom of the element to the mass of $\frac{1}{12}$ part of an atom of $\mathrm{C}-12$.
Relative atomic mass $=\frac{\text { Mass of one atom of the element }}{\text { Mass of } 1 / 12 \text { part of atom of C-12 }}$
Average atomic mass $=\frac{\sum A _{i} X _{i}}{\sum X _{\text {total }}}$
Where $\mathrm{A} _{1}, \mathrm{~A} _{2}, \mathrm{~A} _{3} \ldots \ldots \ldots$ are atomic masses of species $1,2,3 \ldots .$. etc, with their $\%$ or fractions $\mathrm{X} _{1}, \mathrm{X} _{2}, \mathrm{X} _{3}$ etc.
Similar terms are for molar masses.
Calculation of atomic weight
Atomic wt. $\times$ specific heat $=6.4 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ (approx.) (Dulong and Petit’s law for solids)
Approx. atomic weight $=6.4 /$ specific heat
Valency $=\frac{\text { Atomic mass }}{\text { Equivalent mass }}$
Atomic mass $=$ Eq. mass $x$ Valency
Atomic Mass Unit, u (earlier called atomic mass unit, amu) is equal to $1 / 12$ th of the mass of one ${ }^{12} \mathrm{C}$ atom
$1 \mathrm{u}=\frac{1}{12} \times \frac{12 \mathrm{~g} \mathrm{~mol}^{-1}}{6.022 \times 10^{23} \mathrm{~mol}^{-1}}$
$=1.66 \times 10^{-24} \mathrm{~g}$
Most of the elements have isotopes, the atomic mass of an element is the weighted average of the masses of its all the naturally occurring isotopes.
$M _{a v}=\frac{m _{1} x a+m _{2} x b+m _{3} x c}{a+b+c}$
$m _{1}, m _{2}$ and $m _{3}$ are the atomic masses of the isotopic forms of an element in the ratio $a, b$ and $\mathrm{c}$.
$\mathrm{M} _{\mathrm{av}}$ is the average atomic mass
If relative abundance is given instead of ratio
$M _{a v}=\frac{m _{1} x r _{1}+m _{2} x r _{2}+m _{3} x r _{3}}{r _{1}+r _{2}+r _{3}}$
$r _{1}, r _{2}$ and $r _{3}$ are the relative abundances of the isotopes
Molar mass:
Molar mass of a substance is the mass of one mole of that substance.
Formula mass:
Formula mass of a substance is the sum of the atomic masses of all atoms present in one formula unit of the compound (normally ionic compounds)
Gram-molar volume :
The volume occupied by one gram-molecular mass of any gas at NTP $\left(0^{\circ} \mathrm{C}\right.$ or $273 \mathrm{~K}$ and one atm or $760 \mathrm{~mm}$ of $\mathrm{Hg}$ pressure). Its value is 22.4 litre. It is also known as molar volume.
Vapour density :
V.D. $=\frac{\text { Density of a gas }}{\text { Density of hydrogen }}$
$=\frac{\text { Mass of a certain volume of a gas }}{\text { Mass of same volume of hydrogen }}$
$2 \times$ V.D. $=$ Molecular mass
Solved Examples
Question 1. Convert the $28^{\circ} \mathrm{C}$ temperature into degrees Fahrenheit heat
b) $81.4^{\circ}$
a) $77.0^{\circ}$
c) $77.4^{\circ}$
d) 81.0
Show Answer
Answer : (b)
$\begin{aligned} & { }^{\circ} \mathrm{F}=\frac{9\left({ }^{\circ} \mathrm{C}\right)}{5}+32 \\ & =\frac{9}{5} \times 28+32=81.4^{\circ} \end{aligned}$
Question 2. What will be the answer of $77.32-6.3$ ?
a) 71.02
b) 71.0
c) 71.2
d) 71.1
Show Answer
Answer : (b)
The answer is to be reported upto same number of decimal places as that of the term with the least number of decimal places. This after rounding off 71.02 becomes 71.0 .
Question 3. A student performs a titration with different burettes and finds titre values of $25.2 \mathrm{~mL}$, $25.25 \mathrm{~mL}$ and $25.0 \mathrm{~mL}$ the number of significant figures in the average titre value is
a) 1
b) 3
c) 2
d) 4
Show Answer
Answer : (b)
Calculate the average value of different titre values. Then use the rule to determine the significant figures for mathematical calculation.
Average value $=\frac{25.2+25.25+25.0}{3}=\frac{75.45}{3}$
$=25.15$
$=25.2 \mathrm{~mL}$
The number of significant figure is 3
Practice Questions
1. A gaseous mixture contains oxygen and nitrogen in the ratio of $1: 4$ by weight. Therefore, the ratio of their number of molecules is
(a) $1: 4$
(b) $1: 8$
(c) $7: 32$
(d) $3: 16$
Show Answer
Answer: (c)2. The total number of electrons in one molecule of carbon dioxide is
(a) 22
(b) 44
(c) 66
(d) 88
Show Answer
Answer: (a)3. The largest number of molecules is in
(a) $36 \mathrm{~g}$ of water
(b) $28 \mathrm{~g}$ of $\mathrm{CO}$
(c) $46 \mathrm{~g}$ of ethylalcohol
(d) $54 \mathrm{~g}$ of nitrogen pentaoxide $\left(\mathrm{N} _{2} \mathrm{O} _{5}\right)$
Show Answer
Answer: (a)4. When the same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide, the ratio of volumes of hydrogen evolved is
(a) $1: 1$
(b) $1: 2$
(c) $2: 1$
(d) $9: 4$
Show Answer
Answer: (a)5. $\quad 2.76 \mathrm{~g}$ of silver carbonate on being strongly heating yields a residue weighing
(a) $2.16 \mathrm{~g}$
(b) $2.48 \mathrm{~g}$
(c) $2.32 \mathrm{~g}$
(d) $2.64 \mathrm{~g}$
Show Answer
Answer: (a)6. Which has maximum number of molecules?
(a) $7 \mathrm{~g}$ of $\mathrm{N} _{2}$
(b) $2 \mathrm{~g}$ of $\mathrm{H} _{2}$
(c) $16 \mathrm{~g}$ of NO$ _{2}$
(d) $16 \mathrm{~g} _{\text {of }}^{2}$
Show Answer
Answer: (b)7. $2.6 \mathrm{~g}$ of a mixture of calcium carbonate and magnesium carbonate is strongly heated to constant mass $1.3 \mathrm{~g}$. The atomic weights of calcium and magnesium are 40 and 24 respectively. State which of the following masses expresses the mass of calcium carbonate in the original mixture?
(a) $980 \mathrm{mg}$
(b) $400 \mathrm{mg}$
(c) $1.75 \mathrm{~g}$
(d) $0.74 \mathrm{~g}$
Show Answer
Answer: (d)8. If $0.44 \mathrm{~g}$ of a colourless oxide of nitrogen occupies $224 \mathrm{~mL}$ at $1520 \mathrm{~mm} \mathrm{Hg}$ and $273^{\circ} \mathrm{C}$, then the compound is
(a) $\mathrm{N} _{2} \mathrm{O}$
(b) $\mathrm{NO} _{2}$
(c) $\mathrm{NO} _{4}$
(d) $\mathrm{N} _{2} \mathrm{O} _{2}$
Show Answer
Answer: (b)9. If oxygen is present in one litre flask at pressure of $7.6 \times 10^{-10} \mathrm{~mm} \mathrm{Hg}$, then the number of oxygen molecules in the flask at $0^{\circ} \mathrm{C}$ will be
(a) $27 \times 10^{10}$
(b) $ 0.27 \times 10^{10}$
(c) $0.027 \times 10^{10}$
(d) $2.7 \times 10^{10}$
Show Answer
Answer: (a)10. The ratio of masses of oxygen and nitrogen and a particular gaseous mixture is $1: 4$. The ratio of number of their molecule is
(a) $1: 4$
(b) $7: 32$
(c) $1: 8$
(d) $3: 16$
Show Answer
Answer: (b)Mole
The quantity of a given substance that contains as many elementary entities, such as atoms, molecules ions or formula units as the number of atoms in exactly $12 \mathrm{~g}$ of C-12.
In modern terms, gram-molecule and gram-atom are termed as a mole of molecules and a mole of atoms respectively e.g., 1 gram-molecule of oxygen and I gram-atom of oxygen are expressed as 1 mole of $\mathrm{O} _{2}$ and 1 mole of $\mathrm{O}$ respectively.
1 Mole of atoms $=$ Gram atomic mass or molar mass $=6.022 \times 10^{23}$ molecules $=22.4 \mathrm{~L}$ at NTP $\left(0^{\circ} \mathrm{C}, 1\right.$ $\mathrm{atm})=22.7 \mathrm{~L}$ at STP $\left(0^{\circ} \mathrm{C}, 1\right.$ bar $)$
The number of moles of a substance can be calculated by various means.
Rules in Brief
(1) $\quad$ Number of moles of molecules $=\frac{\text { Mass ing }}{\text { Molar mass }}$
(2) $\quad$ Number of moles of atoms $=\frac{\text { Mass ing }}{\text { Molar mass }}$
(3) $\quad$ Number of moles of gases $=\frac{\text { Volume at NTP }}{\text { Standard molar volume }}$
(Standard molar volume is the volume occupied by 1 mole of any gas at NTP which is equal to 22.4 litres.)
(4) Number of moles of atoms/molecules/ions/electrons
$=\frac{\text { No. of atoms } / \text { molecules/ions/electrons }}{\text { Avogadro constant }}$
(5) Number of moles of solute $=$ Molarity $x$ Volume of solution in litres
Percentage Composition
$\%$ by weight of solute in solution $=\frac{\text { mass of solute in gram }}{\text { mass of solution in gram }} \times 100$
$\%$ by volume of solute in solution $=\frac{\text { volume of solute in } \mathrm{mL}}{\text { volume of solution in } \mathrm{mL}} \times 100$
$\%$ by weight to volume of solute in solution $=\frac{\text { mass of solute in gram }}{\text { volume of solution in } \mathrm{mL}} \times 100$
Empirical and Molecular formula:
Steps involved in the calculation of empirical formula:
Convert the mass percentage into grams: considering $100 \mathrm{~g}$ of the compound, the given mass percentages represent the asses of the elements in grams.
Calculate the number of moles of each element.
Calculate the simplest molar ratio: divide the moles obtained by the least value from amongst the values obtained for each element.
Calculate the simplest whole number ratio: multiply all the simplest atomic ratios by a suitable integer.
Write the empirical formula: write the symbols of each element along with their whole number ratios side by side.
Molecular formula = empirical formulaxn
$n=1,2,3, \ldots$. etc.
$\mathrm{n}=\frac{\text { molar mass }}{\text { empirical formula mass }}$
Methods of expressing concentration of a solution
The amount of solute in a solution can be expressed in terms of two systems of units :
(i) Group A units specify the amount of solute in a given volume of solution, such as strength molarity, normality etc.
(ii) Group B units specify the amount of solute for a given mass of solvent or solution, such as mass %, mole fraction, molality etc.
The advantage of Group A units is the ease of solution preparation and that of Group B units is the temperature independence.
In a given solution, let solute be represented by (1) and the solvent by (2)
$\mathrm{w} _{1}=$ mass of solute,
$\mathrm{W} _{2}=$ mass of solvent,
$m _{1}=$ molar mass of solute,
$\mathrm{m} _{2}=$ molar mass of solvent, $\mathrm{n} _{1}=$ moles of solute
$\mathrm{w} _{1}+\mathrm{w} _{2}=$ mass of solution,
$\mathrm{d}=$ density in $\mathrm{g} \mathrm{cm}^{-3}$
$E _{1}=$ equivalent mass of solute $\mathrm{n} _{2}=$ moles of solvent
$\left(\mathrm{n} _{1}+\mathrm{n} _{2}\right)=$ moles of solution,
$\mathrm{V}=$ volume of solution in $\mathrm{cm}^{3}$ or $\mathrm{mL}$
Expressing concentration of solution in Groups $A$ and $B$ units :
Following are the terms given in Table 1 and Table 2 widely used to express concentration of solution.
Table 1 : Expressing Concentrations in Group A
| Name | Symbol | Definition | Formula | Usual SI unit |
|---|---|---|---|---|
| Mass percent | $\mathrm{C}$ | Mass of solute present in $100 \mathrm{~mL}$ of solution Strength | $\mathrm{C}=\frac{100 \mathrm{w} _{1}}{\mathrm{~V}}$ or | $\mathrm{kg} \mathrm{m}^{-3}$ |
| Molarity | $\mathrm{M}$ | Mass of solute present $1 \mathrm{~L}$ or $1 \mathrm{dm}^{3}$ of solution Number of mol of solute present in one $\mathrm{L}\left(1 \mathrm{dm}^{3} /\right.$ $\left.1000 \mathrm{~cm}^{3}\right)$ of solution | $\mathrm{M}=\frac{1000 \mathrm{w} _{1}}{\mathrm{~V}}$ | $\mathrm{~g} \mathrm{~L}^{-1}$ or g dm $\mathrm{m} _{1}^{-3}$ |
Table 2 : Expressing Concentrations in Group B
| Name | Symbol | Definition | Formula | Usual SI unit |
|---|---|---|---|---|
| concentration by mass of solution | $\%$ | Mass of solute present in $100 \mathrm{~g}$ of solution by mass of solvent | $=\frac{100 \mathrm{w} _{1}}{\mathrm{~W} _{1}+\mathrm{w} _{2}}$ | Dimension- |
| less | ||||
| Molality | $\%$ | Mass of solute present in $100 \mathrm{~g}$ of solvent Mumber of mol of solute present in one kg solvent | $\mathrm{M}^{\prime}=\frac{1000 \mathrm{w} _{1}}{\mathrm{w} _{2} \mathrm{~m} _{1}}$ | mol kg $\mathrm{w}^{-1}$ |
| Fraction of a component out of total mol present in a solution | $\mathrm{X} _{1}=\frac{\mathrm{n} _{1}}{\mathrm{n} _{1}+\mathrm{n} _{2}}$ | Dimension- less |
Relationship between molality (M’) and Molarity (M)
$$ \begin{aligned} M^{\prime} & =\frac{1000 M}{\left(1000 d-M _{1}\right)} \\ \frac{d}{M} & =\frac{m _{1}}{1000}+\frac{1}{M^{\prime}} \end{aligned} $$
Relationship between molality ( $\mathrm{M}^{\prime}$ ) and mole fraction $(\mathrm{X})$
$$ \mathrm{M}^{\prime}=\frac{1000 \mathrm{x} _{1}}{\mathrm{x} _{2} \mathrm{~m} _{2}} $$
$$ \frac{d}{M}=\frac{m _{1}}{1000}+\frac{1}{M} $$
Relationship between molarity ( $\mathrm{M}$ ) and Mole fraction $(\mathrm{X})$
$$ M=\frac{1000 w _{2} X _{1} d}{m _{2}\left(w _{1}+w _{2}\right)\left(1-X _{1}\right)} $$
While diluting a solution from one concentration to another, we can use
$$ \begin{aligned} \mathrm{M} _{1} \mathrm{~V} _{1}= & \mathrm{M} _{2} \mathrm{~V} _{2} \\ \text { (Conentrated solution) } & \text { (dilute solution) } \end{aligned} $$
Molarity of mixture :
Let there be three samples of solution (containing same solvent and solute) with their molarity $\mathrm{M} _{1}, \mathrm{M} _{2}, \mathrm{M} _{3}$ and volumes $\mathrm{V} _{1}, \mathrm{~V} _{2}, \mathrm{~V} _{3}$ respectively. These solutions are mixed, molarity of mixed solution may be given as :
$$ \begin{aligned} & M _{1} V _{1}+M _{2} V _{2}+M _{3} V _{3}=M\left(V _{1}+V _{2}+V _{3}\right) \\ & \text { where, } M=\text { resultant molarity } \end{aligned} $$
Equivalent weight
The number of parts by weight of the substance which combine or displace directly or indirectly 1.008 parts by weight of hydrogen or 8 parts by weight of oxygen or 35.5 parts by weight of chlorine or 108 parts by weight of silver.
Methods to calculate equivalent weight
Hydrogen displacement method
$$ \begin{aligned} \text { Eq. wt. of metal } & =\frac{\text { Wt. of metal }}{\text { Wt. of } \mathrm{H} _{2} \text { displaced }} \times 1.008 \\ & =\frac{\text { Wt. of metal }}{\text { Volume of } \mathrm{H} _{2} \text { in } \mathrm{mL} \text { displaced at NTP }} \times 11200 \end{aligned} $$
Oxide formation or reduction of the oxide method:
$$ \begin{aligned} \text { Eq. wt. of metal } & =\frac{\text { Wt. of metal }}{\text { wt. of oxygen combined }} \times 8 \\ & =\frac{\text { wt. of metal }}{\text { Wt. of } \mathrm{O} _{2} \text { displaced/combined in } \mathrm{mL} \text { at NTP }} \times 5600 \end{aligned} $$
Chloride formation method:
$$ \begin{aligned} \text { Eq. wt. of metal } & =\frac{\text { Wt. of metal }}{\text { Wt. of chlorine combined }} \times 35.5 \\ & =\frac{\text { Wt. of metal }}{\text { Wt. of } \mathrm{Cl} _{2} \text { combined in } \mathrm{mL} \text { at NTP }} \times 11200 \end{aligned} $$
Metal displacement method:
$$ \frac{\text { Wt. of metal added to a salt solution }}{\text { Wt. of metal displaced }}=\frac{\text { Eq. Wt. of metal added }}{\text { Eq. Wt. of metal displaced }} $$
Double decomposition method:
For a reaction $\mathrm{AB}+\mathrm{CD} \rightarrow \mathrm{AD} \downarrow+\mathrm{BC}$ $$ \frac{\text { Wt. of salt } A B \text { added to salt } C D \text { (in solution) }}{\text { Wt. of salt } A D \text { precipitated }}=\frac{E q \text {. Wt. of radical } A+E q \text {. wt. of radical } B}{\text { Eq. Wt. of radical } A+E q \text {. wt. of radical } D} $$
Electrolytic method:
Eq. $w t=w t$ deposited by 1 Faraday ( 96500 coulombs)
On passing the same quantity of electricity through two different electrolytic solutions.
$$ \frac{\text { Wt. of } X \text { deposited }}{\text { Wt. of } Y \text { deposited }}=\frac{\text { Eq. wt. of } X}{\text { Eq. wt. of } Y} $$
Neutralization method:
Eq. $w$. of an acid $=$ Wt. of the acid neutralized by $1000 \mathrm{cc}$ of $1 \mathrm{~N}$ base solution
Eq. wt. of a base $=$ Wt. of the base neutralized by $1000 \mathrm{cc}$ of $1 \mathrm{~N}$ acid solution
For an organic acid (RCOOH)
$=\frac{\text { Eq. wt. of silver salt }(\mathrm{RCOOAg})}{\text { Eq. Wt. of Silver (108) }}=\frac{\text { Wt. of silver salt }}{\text { Wt. of silver }}$
Eq. wt. of acid (RCOOH) = Eq. wt. of RCOOAg - 107
Eq.wt. of an acid $=\frac{\text { Mol. Wt. of the acid }}{\text { Basicity }}$
Eq. wt. of base $=\frac{\text { Mol. Wt. of the base }}{\text { Acidity }}$
Eq.wt. of salt $=\frac{\text { Mol. Wt. of the salt }}{\text { Total positive valency of the metal atoms }}$
Eq.wt. of oxidizing/reducing agent $=$ Mol. Wt. of the substance
No. of electrons gained/lost by one molecule
Normality
It is defined as no. of equivalents of a solute present in one litre of solution
$$ \begin{aligned} & N=\frac{\text { Equivalent of solute }}{\text { Volume of solution in litre }} \\ & =\frac{\text { Weight of solute }}{\text { Equivalent weight of solute } x V \text { in litre }} \end{aligned} $$
$$ \begin{aligned} & N=\frac{W}{E x V \text { in }(L)} \\ & N=\frac{W \times 1000}{E x V \text { in mL }} \end{aligned} $$
Also, no. of equivalents $=N x V($ in $L)=\frac{\text { Wt. of solute }}{\text { Eq. wt. of solute }}$
And no. of Milli equivalents $=\mathrm{NxV}($ in $\mathrm{mL})=\frac{\text { Wt. of solute }}{\text { Eq. wt. of solute }} \times 1000$
1. Equivalents and Meq. (milliequivalents) of reactants react in equal amount to give same no. of equivalent or Meq. of products separately.
(i) In a given reaction.
$a A+b B \rightarrow m M+n N$
Meq. of $A=$ Meq. of $B=$ Meq. of $M=$ Meq. of $N$
(ii) In a compound $\mathrm{M} _{\mathrm{x}} \mathrm{N} _{\mathrm{y}}$
Meq. of $\mathrm{M} _{\mathrm{x}} \mathrm{N} _{\mathrm{y}}=$ Meq. of $\mathrm{M}=$ Meq. of $\mathrm{N}$
or Eq. of $\mathrm{M} _{\mathrm{x}} \mathrm{N} _{\mathrm{y}}=\mathrm{Eq}$. of $\mathrm{M}=\mathrm{Eq}$. of $\mathrm{N}$
(iii) In a series of reaction for complete reaction
$\mathrm{aA}+\mathrm{bB} \rightarrow \mathrm{cC} \underset{\text { excess }}{\mathrm{dD}} \mathrm{eE} \xrightarrow[\text { excess }]{\mathrm{fF}} \mathrm{gG}$
Meq. of $A=$ Meq. of $B=$ Meq. of $C=$ Meq. of $D=$ Meq. of $E=$ Meq. of $F=$ Meq. of $G$ used used formed used formed used formed
2. Mole and millimole react according to balanced chemical equation.
3. Molarity $x$ Valency factor $=$ Normality
4. On diluting a solution, mole, mM, Equivalents and Meq. of solute do not change.
5. For reporting concentration of $\mathrm{H} _{2} \mathrm{O} _{2}$, direct conversions can be made as:
(i) % strength of $\mathrm{H} _{2} \mathrm{O} _{2}=\frac{17}{50} \times$ Volume strength of $\mathrm{H} _{2} \mathrm{O} _{2}$
(ii) Volume strength of $\mathrm{H} _{2} \mathrm{O} _{2}=5.6 \times$ Normality of $\mathrm{H} _{2} \mathrm{O} _{2}$
(iii) Volume strength of $\mathrm{H} _{2} \mathrm{O} _{2}=11.2 \times$ Molarity of $\mathrm{H} _{2} \mathrm{O} _{2}$
Balancing of a chemical equation
According to the law of conservations of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation.
Step I Write down the correct formulas of reactants and products.
Step II Balance the number of atoms of elements other than $\mathrm{H} \& 0$.
Step III Balance the number of $\mathrm{H}$ and $\mathrm{O}$ if any.
Step IV Verify the number of atoms for all reactants and products.
Solved Examples
Question 1. How many moles of magnesium phosphate, $\mathrm{Mg} _{3}\left(\mathrm{PO} _{4}\right) _{2}$ will contain 0.25 mole of oxygen atoms?
(a) $1.25 \times 10^{-2}$
(b) $2.5 \times 10^{-2}$
(c) 0.02
(d) $3.125 \times 10^{-2}$
Show Answer
Strategy : using the mole concept equate the moles of $\mathrm{Mg} _{3}\left(\mathrm{PO} _{4}\right) _{2}$ to the number of atoms of oxygen in the formula of the compound.
Solution : 1 mole of $\mathrm{Mg} _{3}\left(\mathrm{PO} _{4}\right) _{2}$ contains 8 moles of oxygen atoms.
Therefore 0.25 moles of oxygen atoms are present in $1 / 8 \times 0.25$ moles of $\mathrm{Mg} _{3}\left(\mathrm{PO} _{4}\right) _{2}$ $=3.125 \times 10^{-2}$ moles of $\mathrm{Mg} _{3}\left(\mathrm{PO} _{4}\right) _{2}$
Answer : (d)
Question 2. In the reaction
$2 \mathrm{Al}($ aq. $)+6 \mathrm{HCl}($ aq. $) \longrightarrow 2 \mathrm{Al}^{3+}($ aq. $)+6 \mathrm{Cl}($ aq. $)+3 \mathrm{H} _{2}(\mathrm{~g})$
(a) $33.6 \mathrm{~L} \mathrm{H} \mathrm{H} _{2}(\mathrm{~g})$ is produced regardless of temperature and pressure for every mole of Al that reacts.
(b) $67.2 \mathrm{~L} \mathrm{H} _{2}(\mathrm{~g})$ at NTP $\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)$ is produced for every mole Al that reacts.
(c) $11.2 \mathrm{~L} \mathrm{H} _{2}(\mathrm{~g})$ at NTP $\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)$ is produced for every mole $\mathrm{HCl}(\mathrm{aq})$ consumed.
(d) $6 \mathrm{~L} \mathrm{HCl}\left(\right.$ aq.) is consumed for every $3 \mathrm{LH} _{2}(\mathrm{~g})$ produced.
Show Answer
Strategy : use the mole concept where moles of $\mathrm{HCl}$ are equated to the volume occupied by $\mathrm{H} _{2}$ gas at N.T.P. as given by the balanced chemical equation.
One mole of a gas at N.T.P. $\left(0^{\circ} \mathrm{C}, 1 \mathrm{~atm}\right)$ occupies $22.4 \mathrm{~L}$ of volume.
Solution : 6 moles of $\mathrm{HCl}$ liberate 3 moles of $\mathrm{H} _{2}$ gas.
Convert the moles of $\mathrm{H} _{2}$ gas to the volume occupied by $\mathrm{H} _{2}$ gas $=3 \times 22.4 \mathrm{~L}^{2}$ of $\mathrm{H} _{2}$
6 moles of $\mathrm{HCl}$ liberate $3 \times 22.4 \mathrm{~L}$ of $\mathrm{H} _{2}$
1 mole of $\mathrm{HCl}$ will liberate $\frac{3 \times 22.4}{6}=11.2 \mathrm{~L}$
Answer : (c)
Practice Questions
Question 1. $4.0 \mathrm{~g}$ of caustic soda (mol. mass 40 ) contains same number of sodium ions as per present in:
(a) $10.6 \mathrm{~g}$ of $\mathrm{Na} _{2} \mathrm{CO} _{3}$ (mol. mass 106)
(b) $58.5 \mathrm{~g}$ of NaCl (formula mass 58.5)
(c) $100 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{Na} _{2} \mathrm{SO} _{4}$ (formula mass 14)
(d) $1 \mathrm{~g}$ equivalent of $\mathrm{NaNO} _{3}$ (mol. mass 85)
Show Answer
Answer : (c)Question 2. $0.1 \mathrm{~g}$ of a metal on reaction with dil acid gave $34.2 \mathrm{~mL}$ hydrogen gas at N.T.P. The equivalent weight of the metal is :
(a) 32.7
(b) 48.6
(c) 64.2
(d) 16.3
Show Answer
Answer : (a)Question 3. $0.5 \mathrm{~g}$ of a metal on oxidation gave $0.79 \mathrm{~g}$ of its oxide. The equivalent weight of the metal is :
(a) 10
(b) 14
(c) 20
(d) 40
Show Answer
Answer : (b)Question 4. $74.5 \mathrm{~g}$ of a metal chloride contains $35.5 \mathrm{~g}$ of chlorine. The equivalent weight of the metal is:
(a) 19.5
(b) 35.5
(c) 39.0
(d) 78.0
Show Answer
Answer : (c)Question 5. The chloride of a metal $(\mathrm{M})$ contains $65.5 $ of chlorine. $100 \mathrm{~mL}$. of the vapour of the chloride of the metal at N.T.P. weights $0.72 \mathrm{~g}$. The molecular formula of the metal chloride is:
(a) $\mathrm{MCl}$
(b) $\mathrm{MCl} _{2}$
(c) $\mathrm{MCl} _{3}$
(d) $\mathrm{MCl} _{4}$
Show Answer
Answer : (c)Question 6. The equivalent weight of a metal is 4.5 and the molecular weight of its chloride is 80 . The atomic weight of the metal is:
(a) 18
(b) 9
(c) 4.5
(d) 36
Show Answer
Answer : (b)Question 7. The sulphate of an element contains $42.2 $ element. The equivalent weight would be :
(a) 17.0
(b) 35.0
(c) 51.0
(d) 68.0
Show Answer
Answer : (b)Question 8. The equivalent weight of an element is 4. Its chloride has a vapour density 59.25. Then the valency of the element is :
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer
Answer : (b)Question 9. The oxide of an element possesses the formula $\mathrm{M} _{2} \mathrm{O} _{3}$. If the equivalent weight of the metal is 9 , then the atomic weight of the metal will be :
(a) 9
(b) 18
(c) 27
(d) none of these
Show Answer
Answer : (c)Question 10. Approximate atomic weight of an element is 29.89 . If its eq. wt. is 8.9 , the exact atomic wt. is :
(a) 26.89
(b) 8.9
(c) 17.8
(d) 26.7
Show Answer
Answer : (d)Question 11. $0.534 \mathrm{~g} \mathrm{Mg}$ displaces $1.415 \mathrm{~g}$ Cu from the salt solution of Cu. Equivalent weight of $\mathrm{Mg}$ is 12. The equivalent weight of Cu would be :
(a) 15.9
(b) 47.7
(c) 31.8
(d) 8.0
Show Answer
Answer : (c)Question 12. The equivalent weight of iron in $\mathrm{Fe} _{2} \mathrm{O} _{3}$ would be :
(a) 18.6
(b) 28
(c) 56
(d) 112.0
Show Answer
Answer : (a)Question 13. $5.3 \mathrm{~g}$ of $\mathrm{Na} _{2} \mathrm{CO} _{3}$ have been dissolved to make $250 \mathrm{~mL}$ of the solution. The normality of the resulting solution will be :
(a) $0.1 \mathrm{~N}$
(b) $0.2 \mathrm{~N}$
(c) $0.4 \mathrm{~N}$
(d) $0.8 \mathrm{~N}$
Show Answer
Answer : (c)Question 14. If $8.3 \mathrm{~mL}$ of a sample of $\mathrm{H} _{2} \mathrm{SO} _{4}(36 \mathrm{~N})$ is diluted by $991.7 \mathrm{~mL}$ of water, the approximate normality of the resulting solution is :
(a) 0.4
(b) 0.2
(c) 0.1
(d) 0.3
Show Answer
Answer : (d)Question 15. $10 \mathrm{~mL}$ of an $\mathrm{HCl}$ solution gave $0.1435 \mathrm{~g}$ of $\mathrm{AgCl}$ when treated with excess of $\mathrm{AgNO} _{3}$. The normality of the resulting solution is :
(a) 0.1
(b) 3
(c) 0.3
(d) 0.2
Show Answer
Answer : (a)Question 16. $500 \mathrm{~mL}$ of a $0.1 \mathrm{~N}$ solution of $\mathrm{AgNO} _{3}$ is added to $500 \mathrm{~mL}$ of a $0.1 \mathrm{~N} \mathrm{KCl}$ solution. The concentration of nitrate in the resulting mixture is :
(a) $0.1 \mathrm{~N}$
(b) $0.05 \mathrm{~N}$
(c) $ 0.01 \mathrm{~N}$
(d) $0.2 \mathrm{~N}$
Show Answer
Answer : (b)Question 17. The ratio of amounts of $\mathrm{H} _{2} \mathrm{~S}$ needed to precipitate all the metal ions from $100 \mathrm{~mL}$ of $1 \mathrm{M}$ $\mathrm{AgNO} _{3}$ and $100 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{CuSO} _{4}$ is :
(a) $1: 2$
(b) $2: 1$
(c) zero
(d) infinite
Show Answer
Answer : (a)Question 18. $1 \mathrm{~L}$ of 18 molar $ _{2} \mathrm{SO} _{4}$ has been diluted to $100 \mathrm{~L}$. The normality of the resulting solution is:
(a) $0.09 \mathrm{~N}$
(b) $0.18 \mathrm{~N}$
(c) $1.80 \mathrm{~N}$
(d) $0.36 \mathrm{~N}$
Show Answer
Answer : (d)Question 19. The normality of $0.3 \mathrm{M}$ phosphorus acid $\left(\mathrm{H} _{3} \mathrm{PO} _{3}\right)$ is :
(a) 0.1
(b) 0.9
(c) 0.3
(d) 0.6
Show Answer
Answer : (d)Question 20. One litre of $\mathrm{N} / 2 \mathrm{HCl}$ solution was heated in a beaker. When volume was reduced to $600 \mathrm{~mL}$, $3.25 \mathrm{~g}$ of HCl was given out. The new normality of solution is :
(a) 6.85
(b) 0.685
(c) 0.1043
(d) 6.50
Show Answer
Answer : (b)Question 21. The hydrated salt $\mathrm{Na} _{2} \mathrm{SO} _{4} \cdot \mathrm{HH} _{2} \mathrm{O}$, undergoes $55 $ loss in weight on heating and becomes anhydrous. The value of $n$ will be:
(a) 5
(b) 3
(c) 7
(d) 10
Show Answer
Answer : (d)Question 22. The total ionic strength (total molarity of all ions) of a solution which is $0.1 \mathrm{M}^{2} \mathrm{CuSO} _{4}$ and $0.1 \mathrm{M} \mathrm{of} \mathrm{Al} _{2}\left(\mathrm{SO} _{4}\right) _{3}$ is :
(a) $0.2 \mathrm{M}$
(b) $0.7 \mathrm{M}$
(c) $0.8 \mathrm{M}$
(d) $1.2 \mathrm{M}$
Show Answer
Answer : (b)Question 23. The isotopic abundance of $\mathrm{C}-12$ and $\mathrm{C}-14$ is $98 $ and $2 $ respectively. What would be the number of $\mathrm{C}-14$ atoms in $12 \mathrm{~g}$ carbon sample :
(a) $1.032 \times 10^{22}$
(b) $3.01 \times 10^{23}$
(c) $5.88 \times 1.0^{23}$
(d) $6.02 \times 10^{23}$
Show Answer
Answer : (a)Question 24. $500 \mathrm{~mL} \mathrm{of}^{2}$ contains $6.00 \times 10^{23}$ molecules at N.T.P. How many molecules are present in $100 \mathrm{~mL}$ of $\mathrm{CO} _{2}$ at N.T.P.?
(a) $6 \times 10^{23}$
(b) $1.5 \times 10^{21}$
(c) $5.88 \times 1.0^{23}$
(d) $1.5 \times 10^{22}$
Show Answer
Answer : (b)Question 25. What will be the volume of the mixture after the reaction?
$\mathrm{NH} _{3}(\mathrm{~g}, 1 \mathrm{~L})+\mathrm{HCl}(\mathrm{g}, 1.5 \mathrm{~L}) \rightarrow \mathrm{NH} _{4} \mathrm{Cl}(\mathrm{s})$
(a) $1.5 \mathrm{~L}$
(b) $0.5 \mathrm{~L}$
(c) $1 \mathrm{~L}$
(d) $0 \mathrm{~L}$
Show Answer
Answer : (b)Question 26. What volume of hydrogen gas at $273 \mathrm{~K}$ and $1 \mathrm{~atm}$ pressure will be consumed in obtaining $21.6 \mathrm{~g}$ of elemental boron (atomic mass $=10.8 \mathrm{~g}$ ) from the reduction of boron trichloride by hydrogen?
(a) $67.2 \mathrm{~L}$
(b) $44.8 \mathrm{~L}$
(c) $22.4 \mathrm{~L}$
(d) $89.6 \mathrm{~L}$
Show Answer
Answer : (a)Question 27. The number of water molecules present in a drop of water (volume $0.0018 \mathrm{~mL}$ ) at room temperature is:
(a) $6.022 \times 10^{19}$
(b) $1.084 \times 10^{18}$
(c) $4.84 \times 10^{17}$
(d) $6.022 \times 10^{23}$
Show Answer
Answer : (a)Question 28. 7.5 grams of a gas occupy 5.6 litres of volume at NTP. The gas is :
(a) NO
(b) $\mathrm{N} _{2} \mathrm{O}$
(c) $\mathrm{CO}$
(d) $\mathrm{CO} _{2}$
Show Answer
Answer : (a)Question 29. For the formation of $3.65 \mathrm{~g}$ of hydrogen chloride gas, what volumes of hydrogen gas and chlorine gas are required at N.T.P. conditions?
(a) 1.12 lit, 1.12 lit
(b) 1.12 lit, 2.24 lit
(c) 3.65 lit, 1.83 lit
(d) 1 lit, 1 lit
Show Answer
Answer : (a)Question 30. How many moles of magnesium phosphate $\mathrm{Mg} _{3}\left(\mathrm{PO} _{4}\right) _{2}$ will contain 0.25 mole of oxygen atoms?
(a) $1.25 \times 10^{-2}$
(b) $2.5 \times 10^{-2}$
(c) 0.02
(d) $3.125 \times 10^{-2}$
Show Answer
Answer : (d)Question 31. Which among the following is the heaviest?
(a) One mole of oxygen
(b) One molecule of sulphur trioxide
(c) $44 \mathrm{~g}$ of carbon dioxide
(d) Ten moles of hydrogen
Show Answer
Answer : (c)Question 32. How many grams of magnesium phosphate, $\mathrm{Mg} _{3}\left(\mathrm{PO} _{4}\right) _{2}$ will contain 0.25 mole of oxygen atoms?
(a) $818.75 \times 10^{-3}$
(b) $818.75 \times 10^{-2}$
(c) $818.75 \times 10^{-5}$
(d) $818.75 \times 10^{-3}$
Show Answer
Answer : (b)Question 33. The molality of $1 \mathrm{~L}$ solution of $93 \mathrm{H} _{2} \mathrm{SO} _{4}(\mathrm{w} / \mathrm{v})$ having density $1.84 \mathrm{~g} / \mathrm{mL}$ is :
(a) $1.043 \mathrm{~m}$
(b) $0.1043 \mathrm{~m}$
(c) $10.43 \mathrm{~m}$
(d) $0.01043 \mathrm{~m}$
Show Answer
Answer : (c)Question 34. $0.7 \mathrm{~g}$ of $\mathrm{NaCO} _{3} \cdot \mathrm{xH} _{2} \mathrm{O}$ is dissolved in $100 \mathrm{~mL}$ water, $20 \mathrm{~mL}$ of which required $19.8 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{HCl}$. The value of $x$ is :
(a) 4
(b) 3
(c) 2
(d) 1
Show Answer
Answer : (c)Question 35. The normality of $10 \mathrm{~mL}$ of a ’ $20 \mathrm{~V}^{\prime} \mathrm{H} _{2} \mathrm{O} _{2}$ is :
(a) 1.79
(b) 3.58
(c) 60.86
(d) 6.086
Show Answer
Answer : (d)Question 36. Vapour density of a volatile substance is 4 . Its molecular weight would be:
(a) 8
(b) 2
(c) 64
(d) 128
Show Answer
Answer : (a)Question 37. $3 \mathrm{~g}$ of an oxide of a metal is converted to chloride completely and it yielded $5 \mathrm{~g}$ of chloride. The equivalent weight of the metal is :
(a) 33.25
(b) 3.325
(c) 12
(d) 20
Show Answer
Answer : (a)Question 38. If 0.50 mole of $\mathrm{BaCl} _{2}$ is mixed with 0.20 mole of $\mathrm{Na} _{3} \mathrm{PO} _{4}$, the maximum number of moles of $\mathrm{Ba} _{3}\left(\mathrm{PO} _{4}\right) _{2}$ that can be formed is :
(a) 0.70
(b) 0.50
(c) 0.20
(d) 0.10
Show Answer
Answer : (d)Question 39. A molar solution is one that contains one mole of solute in:
(a) $1000 \mathrm{~g}$ of solvent
(b) 1.0 L of solvent
(c) $1.0 \mathrm{~L}$ of solution
(d) 22.4 L of solution
Show Answer
Answer : (a)Question 40. In which mode of expression, the concentration of a solution remains independent of temperature?
(a) Molarity
(b) Normality
(c) Formality
(d) Molality
Show Answer
Answer : (d)Question 41. How many moles of electron weighs one kilogram?
(a) $6.022 \times 10^{23}$
(b) $1 \times 10^{23} / 9.108$
(c) $6.022 \times 10^{23} / 9.108$
(d) $1 \times 10^{8} / 9.108 \times 6.023$
Show Answer
Answer : (d)Question 42. Which has maximum number of atoms? (At. wts are given within brackets)
(a) $24 \mathrm{~g}$ of $\mathrm{C}(12)$
(b) $56 \mathrm{~g}$ of Fe (56)
(c) $27 \mathrm{~g}$ of $\mathrm{Al}(27)$
(d) $108 \mathrm{~g}$ of $\mathrm{Ag}(108)$
Show Answer
Answer : (a)Question 43. Dissolving $120 \mathrm{~g}$ of urea (mol. wt. 60) in $1000 \mathrm{~g}$ of water gave a solution of density 1.15 $\mathrm{g} / \mathrm{mL}$. The molarity of the solution is :
(a) $1.78 \mathrm{M}$
(b) $2.00 \mathrm{M}$
(c) $2.05 \mathrm{M}$
(d) $2.22 \mathrm{M}$
Show Answer
Answer : (c)Question 44. The molarity of a solution obtained by mixing $750 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{HCl}$ and $250 \mathrm{~mL}$ of $2.0 \mathrm{M}$ $\mathrm{HCl}$ will be:
(a) $0.875 \mathrm{M}$
(b) $1.00 \mathrm{M}$
(c) $1.75 \mathrm{M}$
(d) $0.0975 \mathrm{M}$
Show Answer
Answer : (a)Question 45. The highest mass corresponds to which of the following :
(a) 1 molecule of $\mathrm{O} _{2}$
(b) $1 \times 10^{-23} \mathrm{~g}$ mole of $\mathrm{O} _{2}$
(c) an $0^{2}$ ion
(d) 1 mole of $\mathrm{O} _{2}$
Show Answer
Answer : (d)Question 46. The chloride of a metal has the formula $\mathrm{MCl} _{3}$. The formula of its phosphate will be:
(a) $\mathrm{M} _{2} \mathrm{PO} _{4}$
(b) $\mathrm{MPO} _{4}$
(c) $\mathrm{M} _{3} \mathrm{PO} _{4}$
(d) $\mathrm{M}\left(\mathrm{PO} _{4}\right) _{2}$
Show Answer
Answer : (b)Question 47. By heating $10 \mathrm{~g} \mathrm{CaCO} _{3}, 5.6 \mathrm{~g} \mathrm{CaO}$ is formed, what is the weight of $\mathrm{CO} _{2}$ obtained in this reaction?
(a) $4.4 \mathrm{~g}$
(b) $44 \mathrm{~g}$
(c) $2.2 \mathrm{~g}$
(d) $22 \mathrm{~g}$
Show Answer
Answer : (a)Question 48. If $32 \mathrm{~g}$ of $\mathrm{O} _{2}$ contains $6.022 \times 10^{23}$ molecules at NTP, then $32 \mathrm{~g}$ of $\mathrm{S}$, under the same conditions, will contain :
(a) $6.022 \times 10^{23} \mathrm{~S}$
(b) $3.011 \times 10^{23} \mathrm{~S}$
(c) $12.044 \times 10^{23} \mathrm{~S}$
(d) $1 \times 10^{23} \mathrm{~S}$
Show Answer
Answer : (a)Question 49. The total number of protons in $10 \mathrm{~g}$ of calcium carbonate is
(a) $1.5057 \times 10^{24}$
(b) $2.0478 \times 10^{24}$
(c) $3.0115 \times 10^{24}$
(d) $4.0956 \times 10^{24}$
Show Answer
Answer : (c)Question 50. How many millilitres ( $\mathrm{mL}$ ) of $1 \mathrm{M} \mathrm{H} _{2} \mathrm{SO} _{4}$ solution is required to neutralize $10 \mathrm{~mL}$ of $1 \mathrm{M}$ $\mathrm{NaOH}$ solution?
(a) 2.5
(b) 5.0
(c) 10.0
(d) 20.0
Show Answer
Answer (b)Significance of Chemical Equations
A chemical equation describes the chemical process both qualitatively and quantitatively. The stoichiometric coefficients in the chemical equation give the quantitative information of the chemical process.
Calculations based on Chemical Equations (Stoichiometry)
Calculation based on chemical equations are known as stoichiometric calculations. It is of four types.
i. Calculations involving mass-mass relationship.
ii. Calculations involving mass-volume relationship.
iii. Calculations involving volume-volume relationship.
iv. Calculations involving mole-mole and mole-mass relationship.
Calculations Involving Mass-Mass Relationship
The following steps are used to solve the problem.
i. Write down balanced molecular equations for the chemical changes.
ii. Write down the number of moles below the formula of each of the reactants and products.
iii. Write down the relative masses of the reactants and products with the help of formula below the respective formula. These shall be the theoretical amounts of reactants and products.
iv. By the application of unitary method, the unknown factor or factors are determined.
Calculations Involving Mass-Volume Relationship
These calculations are based on the facts that I mole or I g molecule of the substance occupies $22.4 \mathrm{~L}$ or $22400 \mathrm{~mL}$ at NTP. The following example shows the mass-volume relationship.
$\mathrm{CaCO}_3+2 \mathrm{HCl}=\mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2$
$1 \mathrm{~mol}$ $\qquad2 \mathrm{~mol}$ $\quad\qquad\qquad\qquad1 \mathrm{~mol}$
$100 \mathrm{~g}$ $\qquad73 \mathrm{~g}$ $\qquad\qquad\qquad\qquad44 \mathrm{~g}$
$\qquad\qquad\qquad\qquad\qquad\qquad22.4 \mathrm{~L} \text { or } 22400 \mathrm{~mL} \text { at NTP }$
Volume of a gas at any temperature and pressure can be converted into mass or vice-versa with the help of the equation: $P V=\frac{w}{M} R T$
Where wis the mass of the gas, $M$ is the molar mass and $\mathrm{R}$ is gas constant.
Calculations Involving Volume-Volume Relationship
These calculations are based on two laws:
i. Avogadro’s law and
ii. Gay-Lussac’s law
For example,
$2 \mathrm{NO}+\quad\mathrm{O} _{2}=\quad2 \mathrm{NO} _{2}$
$2 \text{vol.} \qquad1 \text{vol.} \quad2 \text{vol. (Gay-Lussac’s law)}$
Under similar conditions of temperature and pressure, gases react in simple ratio of their volumes.
$ \begin{array}{rll} 2 \mathrm{NO} + & \mathrm{O}_{2}= & 2 \mathrm{NO} _{2} \\ 2 \mathrm{~mol} & 1 \mathrm{~mol} & 2 \mathrm{~mol}\\ 2 \times 22.4 \text{ litre} & 22.4 \text{ litre} & 2 \times 22.4 \text{ litre (Avogadro’s law)} \end{array} $
Under similar conditions of temperature and pressure, equal moles of gases occupy equal volumes.
Example:
| $2 \mathrm{H} _{2}(\mathrm{~g})$ | $\mathrm{O} _{2}(\mathrm{~g})$ | $\rightarrow 2 \mathrm{H} _{2} \mathrm{O}(\mathrm{g})$ | |
| Mole ratio | 2 | 1 | 2 |
| Mass ratio | $4 \mathrm{~g}$ | $32 \mathrm{~g}$ | $36 \mathrm{~g}$ |
| Volume ratio | $2 \mathrm{vol}$ | $1 \mathrm{vol}$ | $2 \mathrm{vol}$ |
| Volumeat STP | $2 \times 22.4 \mathrm{~L}$ | $22.4 \mathrm{~L}$ | $2 \times 22.4 \mathrm{~L}$ |
| Molecular ratio | $2 \times 6.022 \times 10^{23}$ | $1 \times 6.022 \times 10^{23}$ | $2 \times 6.022 \times 10^{23}$ |
1. Calculations Based on Mass-Mass Relationship
Calculate the weight of iron which will be converted into its oxide $\left(\mathrm{Fe} _{3} \mathrm{O} _{4}\right)$ by the action of $14.4 \mathrm{~g}$ of steam on it.
Solution. The balanced chemical equation is :
$ \begin{array}{rcl} 3 \mathrm{Fe} & + &4 \mathrm{H} _{2} \mathrm{O} \rightarrow \mathrm{Fe} _{3} \mathrm{O} _{4}+4 \mathrm{H} _{2} \\ 3 \mathrm{~mol} & & 4 \mathrm{~mol} \\ 3 \times 56 & & 4 \times 18 \\ =168 \mathrm{~g} & & = 72 \mathrm{~g} \end{array} $
Now $72 \mathrm{~g}$ of steam react with $=168 \mathrm{~g}$ of Fe
$14.4 \mathrm{~g}$ of steam will react with $=\frac{168}{72} \times 14.4$
$$ =33.6 \mathrm{~g} \text { of Fe } $$
2. Based on Mass-Volume Relationship
A sample of lime stone contains $80 \mathrm{CaCO} _{3}$. Calculate the volume of $\mathrm{CO} _{2}$ at NTP obtained by treating $25 \mathrm{~g}$ of this sample with excess of dil $\mathrm{HCl}$.
Show Answer
Solution : Mass of pure $\mathrm{CaCO} _{3}=25 \times 0.8=20 \mathrm{~g}$
$\mathrm{CaCO} _{3}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl} _{2}+\mathrm{H} _{2} \mathrm{O}+\mathrm{CO} _{2}$
$100 \mathrm{~g} \qquad\qquad\qquad\qquad\qquad\qquad\qquad 1 \text{mole} $
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad 22.4 LatNTP$
$100 \mathrm{~g} \mathrm{CaCO} _{3}$ gives $\mathrm{CO} _{2}$ at NTP $=22.4 \mathrm{~L}$
$20 \mathrm{~g} \mathrm{CaCO} _{3}$ give CO ${ } _{2}$ at NTP $=\frac{22.4 \times 20}{100}=4.48 \mathrm{~L}$
3. Based on Volume-Volume Relationship
One litre mixture of $\mathrm{CO}$ and $\mathrm{CO} _{2}$ is passed over red hot coke. Volume of the gaseous mixture becomes 1.4 litre. Calculate the volume of $\mathrm{CO}$ and $\mathrm{CO} _{2}$ in the original mixture. All volumes are taken under identical conditions.
Show Answer
Solution : Suppose Volume of $\mathrm{CO} _{2}=$ a litre, Volume of $\mathrm{CO}=(1-\mathrm{a})$ litre
$\mathrm{CO} _{2}$ reacts with coke (carbon). $\mathrm{CO}$ does not react with carbon
$\mathrm{CO} _{2}+\quad\mathrm{C}=\quad2 \mathrm{CO}$
1 litre $\quad\qquad\qquad 2$ litre
a litre $\quad\qquad\qquad$ 2a litre
Volume of $\mathrm{CO}$ formed $=2$ a litres.
Total volume of the gases $=(1-a)+2 a=1.4$ litre
$\mathrm{a}=0.4$ litre (volume of $\mathrm{CO} _{2}$ )
$1-\mathrm{a}=1-0.4=0.6$ litre (volume of $\mathrm{CO}$ )
Another method to solve objective questions of stoichiometry is POAC method.
4. Principle of Atom Conservation (POAC):
If atoms are conserved, moles of atoms shall also be conserved. This is known as principle of atom conservation. This principle is in fact the basis of the mole concept.
Let us see how this principle works. Consider the unbalanced chemical equation :
$\mathrm{KClO} _{3}(\mathrm{~s}) \longrightarrow \mathrm{KCl}(\mathrm{s})+\mathrm{O} _{2}(\mathrm{~g})$
Apply the principle of atom conservation (POAC) for Katoms.
Moles of K atoms in reactant = Moles of K atoms in products
Or
Moles of $\mathrm{Katoms}$ in $\mathrm{KClO} _{3}=$ moles of $\mathrm{K}$ atoms in $\mathrm{KCl}$.
Now, since 1 molecule of $\mathrm{KClO} _{3}$ contains 1 atom of $\mathrm{K}$ or 1 mol of $\mathrm{KClO} _{3}$ contains 1 mole of $\mathrm{K}$ and similarly 1 mole of $\mathrm{KCl}$ contains 1 mole of $\mathrm{K}$.
Thus, moles of $\mathrm{K}$ atoms in $\mathrm{KClO} _{3}=1 \times$ moles of $\mathrm{KClO} _{3}$
And moles of $\mathrm{K}$ atoms in $\mathrm{KCl}=1 \times$ moles of $\mathrm{KCl}$
Moles of $\mathrm{KClO} _{3}=$ moles of $\mathrm{KCl}$
$\frac{\text { Mass of } \mathrm{KClO} _{3} \text { in } \mathrm{g}}{\text { Molar mass of } \mathrm{KClO} _{3}}=\frac{\text { Mass of } \mathrm{KCl} \text { ing }}{\text { Molar mass of } \mathrm{KCl}}$
The above equation gives the weight relationship between $\mathrm{KClO} _{3}$ and $\mathrm{KCl}$ which is important in stoichiometric calculations.
Again applying principle of atoms conservation of 0 atoms,
Moles of $\mathrm{O}$ in $\mathrm{KClO} _{3}=$ moles of $\mathrm{O}$ in $\mathrm{O} _{2}$.
But since 1 mole of $\mathrm{KClO} _{3}$ contains 3 moles of 0 and 1 mole of $\mathrm{O} _{2}$, contains 2 moles of 0 thus moles of $\mathrm{O}$ in $\mathrm{KClO} _{3}=3$ x moles of $\mathrm{KClO} _{3}$.
Moles of $\mathrm{O}$ in $\mathrm{O} _{2}=2 \mathrm{x}$ moles of $\mathrm{O} _{2}$
Therefore $3 x$ moles of $\mathrm{KClO} _{3}=2 x$ moles of $\mathrm{O} _{2}$
Or $3 \times \frac{\text { Wt. of } \mathrm{KClO} _{3} \text { ing }}{\text { Mol. Wt. of } \mathrm{KClO} _{3}}=2 \times \frac{\text { Volume of } \mathrm{O} _{2} \text { at NTP }}{\text { Standard Molar Volume }}$
The above equation thus gives weight-volume relationship of reactant and product.
Question. A sample of $\mathrm{KClO} _{3}$ on decomposition yielded $448 \mathrm{~mL}$ of oxygen gas at NTP. Calculate
i. Wt. of oxygen produced
ii. Wt. of $\mathrm{KClO} _{3}$ originally taken and
iii. Wt. of $\mathrm{KCl}$ produced
$(\mathrm{K}=39, \mathrm{Cl}=35.5 \text { and } 0=16)$
Show Answer
Solution :
i. Mole of oxygen $=\frac{448}{22400}=0.02$
Wt. of oxygen $=0.02 \times 32=0.64 \mathrm{~g}$
ii. $\mathrm{KClO} _{3} \rightarrow \mathrm{KCl}+\mathrm{O} _{2}$
Apply POAC for 0 atoms,
Moles of $\mathrm{O}$ atoms in $\mathrm{KClO} _{3}=$ moles of $\mathrm{O}$ atoms in $\mathrm{O} _{2}$
$3 \times$ moles of $\mathrm{KClO} _{3}=2 x$ moles of $\mathrm{O} _{2}$
( 1 mole of $\mathrm{KClO} _{3}$ contains 3 moles of 0 and 1 moles of $\mathrm{O} _{2}$ contains 2 moles of 0 )
$3 \times \frac{\text { Wt. of } \mathrm{KClO} _{3} \text { in g }}{\text { Mol. Wt. of } \mathrm{KClO} _{3}}=2 \times \frac{\text { Volumeat NTP (litres) }}{22.4}$
$3 \times \frac{\text { Wt. of } \mathrm{KClO} _{3}}{122.5}=2 \times \frac{0.448}{22.4}$
Wt. of $\mathrm{KClO} _{3}=1.634 \mathrm{~g}$.
iii. Again applying POAC for K atoms
$1 \times$ moles of $\mathrm{KClO} _{3}=1 \times$ moles of $\mathrm{KCl}$
( 1 mole of $\mathrm{KClO} _{3}$ contains 1 mole of $\mathrm{K}$ and 1 mole of $\mathrm{KCl}$ contains 1 mole of $\mathrm{K}$ )
$1 x \frac{\text { Wt. of } \mathrm{KClO} _{3}}{\text { Mol. wt. of } \mathrm{KClO} _{3}}=1 x \frac{\text { Wt. of } \mathrm{KCl}}{\text { Mol. wt. of } \mathrm{KCl}}$
$\frac{1.634}{122.5}=\frac{\text { Wt. of KCl }}{74.5}$
$\Rightarrow$ Wt. of KCl $=0.9937 \mathrm{~g}$
Limiting Reactant- The reactant which is completely consumed in a chemical reaction.
The reactant producing the least number of moles of the product is the limiting reactant.
$ \begin{array}{clclcl} & \mathrm{A} & + &2 \mathrm{~B} & \longrightarrow & 4 \mathrm{C} & \\ \text{Initially} & 5 \text{ mole} & & 12 \text{ mole} & &0 \text{ mole} \\ \text{Finally} & 0 \text{ mole} & & 2 \text{ mole} & & 20 \text{ mole} \\ \end{array} $
Hence $A$ is the limiting reactant.
Or The reactant with the least number of equivalents (or milli equivalants) is the limiting reactant.
Question. $500 \mathrm{~mL}$ of $0.25 \mathrm{M} \mathrm{Na} _{2} \mathrm{SO} _{4}$ is added to an aqueous solution of $15 \mathrm{~g}$ of $\mathrm{BaCl} _{2}$ resulting in the formation of white precipitate of insoluble $\mathrm{BaSO} _{4}$. How many moles and how many grams of $\mathrm{BaSO} _{4}$ are formed?
Show Answer
Solution : Reaction involved in the process is :
$BaCl_2 \quad+ \quad Na SO_4 \quad \rightarrow\quad BaSO_4 \quad + \quad 2 NaCl $
$\text {1 mol \qquad\qquad 1 mol \qquad\qquad 1 mol \quad \qquad 2mol } $
Molarity of $\mathrm{Na} _{2} \mathrm{SO} _{4}$ is 0.25 . Thus, $1000 \mathrm{~mL}$ solution contains 0.25 mole $\mathrm{Na} _{2} \mathrm{SO} _{4}$.
Number of moles of $\mathrm{Na} _{2} \mathrm{SO} _{4}$ in $500 \mathrm{~mL}$ will be 0.125
Moles of $\mathrm{BaCl} _{2}=\frac{\text { Mass }}{\text { Molecular mass }}=\frac{15}{208}=0.0721$
1 mole $\mathrm{Na} _{2} \mathrm{SO} _{4}$ reacts with 1 mole $\mathrm{BaCl} _{2}$
0.125 moles $\mathrm{Na} _{2} \mathrm{SO} _{4}$ will react with 0.125 moles $\mathrm{BaSO} _{4}$
Required moles of $\mathrm{BaCl} _{2}>$ available moles of $\mathrm{BaCl} _{2}$
Thus $\mathrm{BaCl} _{2}$ is limiting reactant. 1 mole $\mathrm{BaCl} _{2}$ will give 1 mole $\mathrm{Na} _{2} \mathrm{SO} _{4}$
$\Rightarrow 0.0721$ moles of $\mathrm{BaCl} _{2}$ will give 0.0721 moles of $\mathrm{BaSO} _{4}$
Mass of $\mathrm{BaSO} _{4}=$ number of moles $x$ molecular mass
$$ \begin{aligned} & =0.072 \times 233 \\ & =16.776 \mathrm{~g} \end{aligned} $$
Neutralization
The term neutralization is used for a reaction between an acid and a base or alkali.
Acid + base $=$ salt + water
For example, $\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H} _{2} \mathrm{O}$
Neutralization is a quantitative reaction.
A general definition is based on Bronsted & Lowry acid - base theory.
$$ \mathrm{HA}+\mathrm{B}^{2+} \longrightarrow \mathrm{A}^{-}+\mathrm{BH}^{2+1} $$
HA represents an acid and $B$ represents a base. $Z$ is an electric charge; negative for an anion, zero, or positive for a cation. When the reaction takes place in water and the base is the hydroxide ion, $\mathrm{OH}$, the reaction can be written as
$$ \mathrm{HA}+\mathrm{OH}^{-} \longrightarrow \mathrm{A}+\mathrm{H} _{2} \mathrm{O} $$
When the acid has been neutralized there are no molecules of $\mathrm{HA}$ (or hydrogen ions produced by dissociation of the molecule) left in the solution. It follows that, in a neutralization reaction, the equivalents of base added must be equal to the equivalents of acid present initially. This stage of the reaction to be the equivalence point.
In all cases : equivalents of an acid = equivalents of a base
When neutralization point is reached, $\mathrm{N} _{1} \mathrm{~V} _{1}=\mathrm{N} _{2} \mathrm{~V} _{2}$
End point is the point of completion of the reaction indicated by suitable indicator. Hence it has additional drop of titrating reagent but, we use
$$ N _{1} V _{1}=N _{2} V _{2} $$
Acid Base Titrations
Acid or acid mixture can be titrated against a suitable base or vice-versa using a suitable indicator. These are summarized in the following Table.
Acid-Base Titrations
| Combination | Suitable Indicator | Colour Change | ||
|---|---|---|---|---|
| In acid | In Alkali | |||
| 1 | Strong acid / strong base | Methyl orange | Red | Yellow |
| Bromothymol blue | Yellow | Blue | ||
| 2 | Strong acid / weak base | Methyl orange | Red | Yellow |
| 3 | Weak acid / strong base | Phenolphthalein | Colourless | Pink |
| 4 | Weak acid / weak base | Not suitable for a titration |
Titration of mixture of bases with two indicators
Every indicator has a working range
| Indicator | pH range | Behaving as |
|---|---|---|
| Phenolphthalein | $8-10$ | Weak organic acid |
| Methyl orange | $3-4.4$ | Weak oraganic base |
Thus methyl orange with lower pH range can indicate complete neutralization of all types of bases.
Extent of reaction of different bases with acid $(\mathrm{HCl})$ using these two indicators is summarized below:
| Phenolphthalein | Methyl Orange | |
|---|---|---|
| $\mathrm{NaOH}$ | $100 $ reaction is indicated $\mathrm{NaOH}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{H} _{2} \mathrm{O}$ |
$100 $ reaction is indicated $\mathrm{NaOH}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{H} _{2} \mathrm{O}$ |
| $\mathrm{Na} _{2} \mathrm{CO} _{3}$ | $50 $ reaction upto $\mathrm{NaHCO} _{3}$ Stage is indicated $\mathrm{Na} _{2} \mathrm{CO} _{3}+\mathrm{HCl} \rightarrow \mathrm{NaHCO} _{3}+\mathrm{NaCl}$ |
$100 $ reaction is indicated $\mathrm{Na} _{2} \mathrm{CO} _{3}+2 \mathrm{HCL} \rightarrow 2 \mathrm{NaCl}+\mathrm{H} _{2} \mathrm{O}+\mathrm{CO} _{2}$ |
| $\mathrm{NaHCO} _{3}$ | No reaction is indicated | $100 $ reaction is indicated $\mathrm{NaHCO} _{3}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H} _{2} \mathrm{O}+\mathrm{CO} _{2}$ |
Suppose volume of given standard acid solution (say $\mathrm{HCl}$ ) required
for complete reaction of $\mathrm{Na} _{2} \mathrm{CO} _{3}=\mathrm{xmL}$
for complete reaction of $\mathrm{NaHCO} _{3}=\mathrm{y} \mathrm{mL}$
for complete reaction of $\mathrm{NaOH}=\mathrm{zmL}$
There may be different combination of mixture of bases. We may opt two methods
Method I: We carry two titrations separately with two different indicators
Method II: We carry single titration but adding second indicator after first end point is reached
Results with Two Indicators
| Method I | Method II | ||||
|---|---|---|---|---|---|
| Mixture | Volume of HCI used with indicator | Volume of HCI used with indicator | |||
| Phenolphthalein | Methyl orange | Phenolphthalein | Methyl orange added after firs end point is reached | ||
| 1 | $\mathrm{NaOH}+\mathrm{Na} _{2} \mathrm{CO} _{3}$ | $z+\frac{x}{2} \\ $ $100 \quad 50 $ | $(x+z)$ $100 $ each | $z+\frac{x}{2}$ $100 \quad 50 $ | $\quad \frac{\mathrm{x}}{2}$ (remaining 50% $\mathrm{Na} _{2} \mathrm{CO} _{3}$ is indicated) |
| 2 | $\mathrm{NaOH}+\mathrm{NaHCO} _{3}$ | $z \quad +\quad 0 \\ 100 \qquad $no reaction | $(y+z)$ $100 $ each | $(z+0)$ | y $\{$ remaining $100 \mathrm{NaHCO} _{3}$ is indicated $\}$ |
| 3 | $\mathrm{Na} _{2} \mathrm{CO} _{3}+\mathrm{NaHCO} _{3}$ | $\frac{x}{2}$ $+0 \\ $ $50 \quad $ no reaction | $(x+y)$ $100 $ each | $\frac{(x+0)}{2}$ | $\frac{(x+y)}{2}$ (remaining 50% of $\mathrm{Na} _{2} \mathrm{CO} _{3}$ and $100 \mathrm{NaHCO} _{3}$ are indicated) |
Solved Examples
Question 1. A solution containing $2.675 \mathrm{~g}$ of $\mathrm{CoCl} _{3} .6 \mathrm{NH} _{3}$ (molar mass $=267.5 \mathrm{~g} / \mathrm{mol}$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $\mathrm{AgNO} _{3}$ to give $4.76 \mathrm{~g}$ of $\mathrm{AgCl}$ (molar mass $=143.5 \mathrm{~g} / \mathrm{mol}$ ). The formula of the complex is (At. Mass of $\mathrm{Ag}=$ $108 \mathrm{u})$
(a) $\left\{\mathrm{CoCl}\left(\mathrm{NH} _{3}\right) _{5}\right\} \mathrm{Cl} _{2}$
(b) $\left\{\mathrm{Co}\left(\mathrm{NH} _{3}\right) _{6}\right\} \mathrm{Cl} _{3}$
(c) $\left\{\mathrm{CoCl} _{2}\left(\mathrm{NH} _{3}\right) _{4}\right\} \mathrm{Cl}$
(d) $\left\{\mathrm{CoCl} _{3}\left(\mathrm{NH} _{3}\right) _{3}\right\}$
Show Answer
Strategy: write a chemical equation of the above process. Calculate the number of moles of $\mathrm{AgCl}$ and the number of moles of $\mathrm{CoCl} _{3}$. $6 \mathrm{NH} _{3}$. Using the stoichiometry, equate the moles calculated.
Solution : $\mathrm{CoCl} _{3} \cdot 6 \mathrm{NH} _{3} \rightarrow \mathrm{xCl}^{-} \xrightarrow{\mathrm{AgNO} _{3}} \times \mathrm{AgCl} \downarrow$
Number of moles of $\mathrm{AgCl}=\frac{4.78}{143.6}$
Number of moles of $\mathrm{CoCl} _{3} .6 \mathrm{NH} _{3}=\frac{2.675}{267.5}$
$n(\mathrm{AgCl})=x . n\left(\mathrm{CoCl} _{3} .6 \mathrm{NH} _{3}\right)$
$\frac{4.78}{143.5}=x \frac{2.675}{267.5}$
$x=3.33 \approx 3$
The complexis $\left[\mathrm{Co}\left(\mathrm{NH} _{3}\right) _{6}\right] \mathrm{Cl} _{3}$
Question 2. For a reaction $A+2 B \rightarrow C$, the amount of $C$ formed by starting the reaction with 5 moles of $A$ and 8 moles of $B$ is
(a) 5 moles
(b) 8 moles
(c) 16 moles
(d) 4 moles
Show Answer
Answer : 1 mole of $A$ reacts with 2 moles of $B \Rightarrow 5$ moles of $A$ reacts with 10 moles of $B$ but we have only 8 moles of $B$ so $B$ is the limiting reagent.
Now 2 moles of $B$ form 1 mole of $C$ and 8 moles of $B$ will form 4 moles of $C$-Hence answer is 4 moles.
Practice Question
Question 1. In the reaction, $4 \mathrm{NH} _{3}+5 \mathrm{O} _{2} \rightarrow 4 \mathrm{NO}+6 \mathrm{H} _{2} \mathrm{O}$, when one mole of ammonia and one mole of oxygen are made to react to completion, then
(a) 1.0 mole of $\mathrm{H} _{2} \mathrm{O}$ is produced
(b) all the oxygen is consumed
(c) 1.5 mole of $\mathrm{NO}$ is formed
(d) all the ammoniais consumed
Show Answer
Answer: (b)Question 2. $5 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO} _{3}\right) _{2}$ is mixed with $10 \mathrm{~mL}$ of $0.02 \mathrm{M} \mathrm{KI}$. The amount of $\mathrm{Pbl} _{2}$ precipitated with be about
(a) $ 10^{-3} \mathrm{~mol}$
(b) $ 10^{-4} \mathrm{~mol}$
(c) $2 \times 10^{-4} \mathrm{~mol}$
(d) $4 \times 10^{-3} \mathrm{~mol}$
Show Answer
Answer: (b)Question 3. A solution containing $\mathrm{Na} _{2} \mathrm{CO} _{3}$ and $\mathrm{NaHCO} _{3}$ is titrated against $0.1 \mathrm{M} \mathrm{HCl}$ solution using methyl orange indicator. At the equivalence point
(a) Both $\mathrm{Na} _{2} \mathrm{CO} _{3}$ and $\mathrm{NaHCO} _{3}$ are completely neutralized
(b) Only $\mathrm{Na} _{2} \mathrm{CO} _{3}$ is wholly neutralized
(c) Only $\mathrm{NaHCO} _{3}$ is wholly neutralized
(d) $\mathrm{Na} _{2} \mathrm{CO} _{3}$ is neutralized upto the stage of $\mathrm{NaHCO} _{3}$
Show Answer
Answer: (a)Question 4. A solution containing $\mathrm{Na} _{2} \mathrm{CO} _{3}$ and $\mathrm{NaHCO} _{3}$ is titrated against $0.1 \mathrm{M} \mathrm{HCl}$ solution using phenolphthalein indicator. At the equivalence point
(a) Both $\mathrm{Na} _{2} \mathrm{CO} _{3}$ and $\mathrm{NaHCO} _{3}$ are completely neutralized
(b) Only $\mathrm{Na} _{2} \mathrm{CO} _{3}$ is wholly neutralized
(c) Only $\mathrm{NaHCO} _{3}$ is wholly neutralized
(d) $\mathrm{Na} _{2} \mathrm{CO} _{3}$ is neutralized upto the stage of $\mathrm{NaHCO} _{3}$
Show Answer
Answer: (d)Question 5. A solution containing $\mathrm{NaOH}$ and $\mathrm{Na} _{2} \mathrm{CO} _{3}$ is titrated against $0.1 \mathrm{M} \mathrm{HCl}$ solution using phenolphthalein indicator. At the equivalence point
(a) Both $\mathrm{NaOH}$ and $\mathrm{Na} _{2} \mathrm{CO} _{3}$ are completely neutralized
(b) Only $\mathrm{NaOH}$ is completely neutralized
(c) Only $\mathrm{Na} _{2} \mathrm{CO} _{3}$ is completely neutralized
(d) $\mathrm{NaOH}$ completely and $\mathrm{Na} _{2} \mathrm{CO} _{3}$ upto the stage of $\mathrm{NaHCO} _{3}$ are neutralized.
Show Answer
Answer: (d)Question 6. A mass of $0.355 \mathrm{~g}$ of the compound $\mathrm{M} _{2} \mathrm{CO} _{3} \cdot \mathrm{H} _{2} \mathrm{O}$ (molar mass of $\mathrm{M}=23 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is dissolved in $100 \mathrm{~mL}$ water and titrated against $0.05 \mathrm{M} \mathrm{HCl}$ using methyl orange indicator. If the volume of $\mathrm{HCl}$ consumed is $100 \mathrm{~mL}$ the value of $\mathrm{x}$ is
(a) 2
(b) 5
(c) 7
(d) 10
Show Answer
Answer: (a)Question 7. $20 \mathrm{~mL}$ of a solution containing $0.34 \mathrm{~g}$ of an impure sample of $\mathrm{H} _{2} \mathrm{O} _{2}$ reacts with $0.465 \mathrm{~g}$ of $\mathrm{KMnO} _{4}$ (molar mass $=155 \mathrm{~g} \mathrm{~mol}^{-1}$ ) in acidic medium. The percent purity of $\mathrm{H} _{2} \mathrm{O} _{2}$ is about
(a) $60 $
(b) $70 $
(c) $75 $
(d) $80 $
Show Answer
Answer: (c)Question 8. A mixture of $\mathrm{Na} _{2} \mathrm{CO} _{3}$ and $\mathrm{NaOH}$ in a solution requires $20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$ solution for neutralization when phenolphthalein indicator is used. The volume consumed is $25 \mathrm{~mL}$ when methyl orange indicator is used. The mass per cent of $\mathrm{Na} _{2} \mathrm{CO} _{3}$ in the given mixture is about
(a) $47 $
(b) $57 $
(c) $67 $
(d) $75 $
Show Answer
Answer: (a)Question 9. The mass of $\mathrm{AgCl}$ precipitated when $4.68 \mathrm{~g}$ of $\mathrm{NaCl}$ is added to a solution containing $6.8 \mathrm{~g}$ of $\mathrm{AgNO} _{3}$ is
(a) $ 4.52 \mathrm{~g}$
(b) $ 5.74 \mathrm{~g}$
(c) $ 7.18 \mathrm{~g}$
(d) $ 8.2 \mathrm{~g}$
Show Answer
Answer: (b)Question 10. Twenty milliliter of a solution is $0.1 \mathrm{M}$ in each of $\mathrm{Na} _{2} \mathrm{CO} _{3}$ and $\mathrm{NaHCO} _{3}$. It is titrated against $0.1 \mathrm{M}$ $\mathrm{HCl}$ using phenolphthalein as the indicator. The volume of $\mathrm{HCl}$ used at the end point will be
(a) $10 \mathrm{~mL}$
(b) $ 20 \mathrm{~mL}$
(c) $ 30 \mathrm{~mL}$
(d) $60 \mathrm{~mL}$
Show Answer
Answer: (b)Question 11. An aqueous solution of $6.3 \mathrm{~g}$ oxalic acid dihydrate is made upto $250 \mathrm{~mL}$. The volume of $0.1 \mathrm{~N}$ $\mathrm{NaOH}$ required to completely neutralize $10 \mathrm{~mL}$ of this solution is
(a) $ 40 \mathrm{~mL}$
(b) $ 20 \mathrm{~mL}$
(c) $ 10 \mathrm{~mL}$
(d) $ 4 \mathrm{~mL}$
Show Answer
Answer: (a)Question 12. Ten milliliter of $0.01 \mathrm{M}$ iodine solution is titrated against $0.01 \mathrm{M}$ sodium thiosulphate solution using starch solution. The volume of sodium thiosulphate consumed upto the end point is
(a) $10 \mathrm{~mL}$
(b) $ 15 \mathrm{~mL}$
(c) $20 \mathrm{~mL}$
(d) $ 30 \mathrm{~mL}$
Show Answer
Answer: (c)Question 13. An aqueous solution of $6.3 \mathrm{~g}$ oxalic acid dihydrate is made up to $250 \mathrm{~mL}$. The volume of $0.1 \mathrm{~N}$ $\mathrm{NaOH}$ required to completely neutralize $10 \mathrm{~mL}$ of this solution is
(a) $40 \mathrm{~mL}$
(b) $20 \mathrm{~mL}$
(c) $10 \mathrm{~mL}$
(d) $4 \mathrm{~mL}$
Show Answer
Answer: (a)Question 14. In the standardization of $\mathrm{Na} _{2} \mathrm{~S} _{2} \mathrm{O} _{3}$ using $\mathrm{K} _{2} \mathrm{Cr} _{2} \mathrm{O} _{7}$ by iodometry, the equivalent mass of $\mathrm{K} _{2} \mathrm{Cr} _{2} \mathrm{O} _{7}$ is
(a) (molar mass)/2
(b) (molar mass)/6
(c) (molar mass)/3
(d) same as molar mass
Show Answer
Answer: (b)Question 15. An aqueous solution of $6.3 \mathrm{~g}$ oxalic acid dihydrate is made upto $250 \mathrm{~mL}$. The volume of $0.1 \mathrm{~N}$ $\mathrm{NaOH}$ required to completely neutralize $10 \mathrm{~mL}$ of this solutions is :
(a) $40 \mathrm{~mL}$
(b) $ 20 \mathrm{~mL}$
(c) $10 \mathrm{~mL}$
(d) $4 \mathrm{~mL}$
Show Answer
Answer: (a)Question 16. No. of oxalic acid molecules in $100 \mathrm{~mL}$ of $0.02 \mathrm{~N}$ oxalic acid are:
(a) $6.022 \times 10^{20}$
(b) $6.022 \times 10^{21}$
(c) $6.022 \times 10^{22}$
(d) $6.022 \times 10^{23}$
Show Answer
Answer: (a)Question 17. $30 \mathrm{~mL}$ of an acid is neutralized by $15 \mathrm{~mL}$ of $0.2 \mathrm{~N}$ alkali. The strength of the acid is
(a) $0.1 \mathrm{~N}$
(b) $ 0.2 \mathrm{~N}$
(c) $ 0.3 \mathrm{~N}$
(d) $0.4 \mathrm{~N}$
Show Answer
Answer: (a)Question 18. To neutralize $100 \mathrm{~mL} 0.1 \mathrm{M} \mathrm{H} _{2} \mathrm{SO} _{4}$, the mass of $\mathrm{NaOH}$ required is :
(a) $40 \mathrm{~g}$
(b) $80 \mathrm{~g}$
(c) $ 0.4 \mathrm{~g}$
(d) $0.8 \mathrm{~g}$
Show Answer
Answer: (d)Question 19. $0.16 \mathrm{~g}$ of a dibasic acid required $25 \mathrm{~mL}$ of decinormal $\mathrm{NaOH}$ solution for complete neutralization. The molecular weight of the acid will be:
(a) 32
(b) 64
(c) 128
(d) 256
Show Answer
Answer: (c)Question 20. The volume of water to be added to $100 \mathrm{~cm}^{3}$ of $0.5 \mathrm{~N} \mathrm{H} _{2} \mathrm{SO} _{4}$ to get decinormal concentration is
(a) $400 \mathrm{~cm}^{3}$
(b) $450 \mathrm{~cm}^{3}$
(c) $500 \mathrm{~cm}^{3}$
(d) $100 \mathrm{~cm}^{3}$
Show Answer
Answer: (a)Question 21. The mass of $70 \mathrm{H} _{2} \mathrm{SO} _{4}$ required for neutralization of $1 \mathrm{~mol}$ of $\mathrm{NaOH}$ is
(a) $ 49 \mathrm{~g}$
(b) $ 98 \mathrm{~g}$
(c) $ 70 \mathrm{~g}$
(d) $ 34 \mathrm{~g}$
Show Answer
Answer: (c)Question 22. If LPG cylinder contains a mixture of butane and isobutene, then the amount of oxygen that would be required for combustion of $1 \mathrm{~kg}$ of it will be
(a) $ 1.8 \mathrm{~kg}$
(b) $ 2.7 \mathrm{~kg}$
(c) $ 4.5 \mathrm{~kg}$
(d) $ 3.58 \mathrm{~kg}$
Show Answer
Answer: (d)Question 23. $12 \mathrm{~L}$ of $\mathrm{H} _{2}$ and 11.2 litres of $\mathrm{Cl} _{2}$ are mixed and exploded. The composition by volume of mixture is
(a) $24 \mathrm{~L}$ of $\mathrm{HCl}$
(b) $ 0.8 \mathrm{LCl} _{2}$ and $20.8 \mathrm{~L} \mathrm{HCl}$
(c) $ 0.8 \mathrm{~L} \mathrm{H} _{2}, 22.4 \mathrm{~L} \mathrm{HCl}$
(d) $ 22.4 \mathrm{~L} \mathrm{HCl}$
Show Answer
Answer: (c)Question 24. If potassium chlorate is $80 $ pure, then $48 \mathrm{~g}$ of oxygen would be produced from (atomic mass of $\mathrm{K}=30$ )
(a) $ 153.12 \mathrm{~g} \mathrm{of} \mathrm{KClO} _{3}$
(b) $ 122.5 \mathrm{~g}$ of $\mathrm{KClO} _{3}$
(c) $245 \mathrm{~g} \mathrm{of} \mathrm{KClO} _{3}$
(d) $ 98.0 \mathrm{~g} _{\text {of }} \mathrm{KClO} _{3}$
Show Answer
Answer: (a)Question 25. The mass of oxygen that would be required to produce enough $\mathrm{CO}$ which completely reduces 1.6 $\mathrm{kg} \mathrm{Fe} _{2} \mathrm{O} _{3}$ (at. mass $\mathrm{Fe}=56$ ) is
(a) $240 \mathrm{~g}$
(b) $ 480 \mathrm{~g}$
(c) $720 \mathrm{~g}$
(d) $960 \mathrm{~g}$
Show Answer
Answer: (b)Question 26. $2.76 \mathrm{~g}$ of silver carbonate on being strongly heated yields a residue weighing
(a) $ 2.16 \mathrm{~g}$
(b) $ 2.48 \mathrm{~g}$
(c) $ 2.32 \mathrm{~g}$
(d) $ 2.64 \mathrm{~g}$
Show Answer
Answer: (a)Question 27. $30 \mathrm{~g}$ of Magnesium and $30 \mathrm{~g}$ of oxygen are reacted, then the residual mixture contains
(a) $60 \mathrm{~g}$ of magnesium oxide only
(b) $40 \mathrm{~g}$ of magnesium oxide and $20 \mathrm{~g}$ of oxygen
(c) $45 \mathrm{~g}$ of magnesium oxide and $15 \mathrm{~g}$ of oxygen
(d) $50 \mathrm{~g}$ of magnesium oxide and $10 \mathrm{~g}$ of oxygen
Show Answer
Answer: (d)Question 28. Minimum volume of $\mathrm{SO} _{2}$ gas at N.T.P. which reduces $100 \mathrm{~mL}$ of $0.1 \mathrm{NK} _{2} \mathrm{Cr} _{2} \mathrm{O} _{7}$ is
(a) $ 11.2 \mathrm{~mL}$
(b) $ 22.4 \mathrm{~mL}$
(c) $40 \mathrm{~mL}$
(d) $112 \mathrm{~mL}$
Show Answer
Answer: (d)Question 29. In Haber process, gaseous nitrogen and hydrogen react to form ammonia whose volume as compared to that of reactants (N.T.P.) would be :
(a) one fourth
(b) one half
(c) the same
(d) three fourth
Show Answer
Answer: (b)Question 30. $4.35 \mathrm{~g}$ of a sample of pyrolusite $\left(\mathrm{MnO} _{2}\right)$ when heated with conc. $\mathrm{HCl}$ gave chlorine. The chlorine when passed through potassium iodide solution liberated $6.35 \mathrm{~g}$ of iodine. The percentage purity of the pyrolusite sample is
(a) 40
(b) 50
(c) 60
(d) 70
Show Answer
Answer: (b)Practice Questions
Subjective Type:
Question 1. How many $\mathrm{mL}$ of $0.1 \mathrm{~N} \mathrm{HCl}$ are required to react completely with $1 \mathrm{~g}$ mixture of $\mathrm{Na} _{2} \mathrm{CO} _{3}$ and $\mathrm{NaHCO} _{3}$ containing equimolar amounts of the two?
Show Answer
Answer: $157.8 \mathrm{ml}$Question 2. $0.5 \mathrm{~g}$ mixture of $\mathrm{K} _{2} \mathrm{CO} _{3}$ and $\mathrm{Li} _{2} \mathrm{CO} _{3}$ required $30 \mathrm{~mL}$ of $0.25 \mathrm{~N} \mathrm{HCl}$ for neutralization. What is the % composition of mixture. ( $\mathrm{K}=39, \mathrm{Li}=7$ )
Show Answer
Answer: $\%$ of $\mathrm{K} _{2} \mathrm{CO} _{3}=96 $
$\qquad \qquad \% $ of $\mathrm{Li} _{2} \mathrm{CO} _{3}=4 $
Question 3. A solution containing $4.2 \mathrm{~g}$ of $\mathrm{KOH}$ and $\mathrm{Ca}(\mathrm{OH}) _{2}$ is neutralized by an acid. If it consumes 0.1 equivalent of acid, calculate composition of sample in solution.
Show Answer
Answer:$\%$ of $\mathrm{KOH}=35 $
$\qquad \qquad $ $\%$ of $\mathrm{Ca}(\mathrm{OH}) _{2}=65 $
Question 4. A solution contains $4 \mathrm{~g}$ of $\mathrm{Na} _{2} \mathrm{CO} _{3}$ and $\mathrm{NaCl}$ in $250 \mathrm{~mL}$. $25 \mathrm{~mL}$ of this solution required $50 \mathrm{~mL}$ of $\mathrm{N} / 10 \mathrm{HCl}$ for complete neutralization. Calculate % composition of mixture.
Show Answer
Answer: $\% $ of $\mathrm{Na} _{2} \mathrm{CO} _{3}=66.25 $
$\qquad \qquad$ $\% $ of $\mathrm{NaCl}=33.75 $
Question 5. What weight of $\mathrm{Na} _{2} \mathrm{CO} _{3}$ of $95 $ purity would be required to neutralize $45.6 \mathrm{~mL}$ of $0.235 \mathrm{~N}$ acid?
Show Answer
Answer: $W=0.5978 \mathrm{~g}$Question 6. Calculate the normality of the resulting solution made by adding $2 \mathrm{drops}(0.1 \mathrm{~mL})$ of $0.1 \mathrm{~N} \mathrm{H} _{2} \mathrm{SO} _{4}$ in 1 litre of distilled water.
Show Answer
Answer: $\mathrm{N}=10^{-5}$Question 7. What weight of $\mathrm{AgCl}$ will be precipitated when a solution containing $4.77 \mathrm{~g} \mathrm{NaCl}$ is added to a solution of $5.77 \mathrm{~g}$ of $\mathrm{AgNO} _{3}$ ?
Show Answer
Answer: $W=4.87 \mathrm{~g}$Question 8. How much $\mathrm{AgCl}$ will be formed by adding $200 \mathrm{~mL}$ of $5 \mathrm{~N} \mathrm{HCl}$ to a solution containing $1.7 \mathrm{~g}$ $\mathrm{AgNO} _{3}$ ?
Show Answer
Answer: $W=1.435 \mathrm{~g}$Question 9. What is the normality and nature of a mixture obtained by mixing $0.62 \mathrm{~g}$ of $\mathrm{Na} _{2} \mathrm{CO} _{3} \cdot \mathrm{H} _{2} \mathrm{O}$ to 100 $\mathrm{mL}$ of $0.1 \mathrm{~N} \mathrm{H} _{2} \mathrm{SO} _{4}$ ?
Show Answer
Answer: 0.1N $\qquad \qquad $ NeutralQuestion 10. $0.5 \mathrm{~g}$ of fuming $\mathrm{H} _{2} \mathrm{SO} _{4}$ is diluted with water the solution requires $26.7 \mathrm{~mL}$ of $0.4 \mathrm{~N} \mathrm{NaOH}$ for its neutralization. Find the % of $\mathrm{free} _{\mathrm{SO} _{3}}$ in the sample of oleum?
Show Answer
Answer: $ \% $ of $\mathrm{SO} _{3}=20.78 $Question 11. A sample of mixture of $\mathrm{CaCl} _{2}$ and $\mathrm{NaCl}$ weighing $4.22 \mathrm{~g}$ was heated to precipitate all the $\mathrm{Ca}^{2+}$ ions as $\mathrm{CaCO} _{3}$ which is then quantitatively converted to $0.929 \mathrm{~g}$ of $\mathrm{CaO}$. Calculate the % of $\mathrm{CaCl} _{2}$ in the mixture.
Show Answer
Answer: $1.84 \mathrm{~g}, 43 $Question 12. $10 \mathrm{~mL}$ of $\mathrm{H} _{2} \mathrm{SO} _{4}$ solution [density $1.84 \mathrm{~g} / \mathrm{cc}$ ] contains $98 \mathrm{H} _{2} \mathrm{SO} _{4}$ by weight. Calculate the volume of $2.5 \mathrm{~N} \mathrm{NaOH}$ solution required to just neutralize the acid.
Show Answer
Answer: $147.2 \mathrm{ml}$Question 13. One gram of a sample of lime stone was treated with $25 \mathrm{~mL}$ of $1 \mathrm{~N} \mathrm{HCl}$ solution and the volume was made upto $250 \mathrm{~mL}$ with water, $25 \mathrm{~mL}$ of this solution required $9 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{NaOH}$ for neutralization. Determine the percentage of $\mathrm{CaCO} _{3}$ in the sample.
Show Answer
Answer: $80 \% $Question 14. If a $20 \mathrm{~mL}$ of $0.5 \mathrm{~N} \mathrm{NaOH}$ solution is mixed with $30 \mathrm{~mL}$ of $0.3 \mathrm{~N} \mathrm{HCl}$, find out whether the solution is acidic or basic? Calculate the normality of the resulting solution with respect to acidic or basic solution.
Show Answer
Answer: $1 / 50 \mathrm{~N}$Question 15. $100 \mathrm{~mL}$ of a solution of sulphuric acid, $50 \mathrm{~mL}, \mathrm{~N} / 10$ solution of $\mathrm{NaOH}$ of density $1.28 \mathrm{~g} / \mathrm{cc}$ was added and the volume made upto $200 \mathrm{~mL}, 20 \mathrm{~mL}$ of this solution required $17 \mathrm{~mL}$ of N/10 $\mathrm{Na} _{2} \mathrm{CO} _{3}$ solution for neutralization. Determine the normality of sulphuric acid solution.
Show Answer
Answer: $0.22 \mathrm{~N}$Question 16. $1.575 \mathrm{~g}$ of oxalic acid $(\mathrm{COOH}) _{2} \times \mathrm{H} _{2} \mathrm{O}$ crystal were dissolved in water and the solution made upto $250 \mathrm{~mL} .20 .85 \mathrm{~mL}$ of this solution required $25 \mathrm{~mL}$ of N/12 NaOH for complete neutralization. Calculate the value of $\mathrm{x}$ ?
Show Answer
Answer: 2Question 17. $0.7324 \mathrm{~g}$ of zinc dust, containing $\mathrm{Zn}$ and $\mathrm{ZnO}$ were dissolved in dil $\mathrm{H} _{2} \mathrm{SO} _{4}$ liberated $224 \mathrm{~mL}$ of $\mathrm{H} _{2}$ at NTP. Calculate the $\%$ of $Z n$ metal present in the sample. $(Z n=65.4)$.
Show Answer
Answer: $89.3 \ %$Question 18. Find out the molarity of a solution obtained by mixing $2.5 \mathrm{~L}$ of $0.1 \mathrm{M} \mathrm{NaOH}, 500 \mathrm{ML}$ water, 0.05 mole $\mathrm{NaOH}$ and $4 \mathrm{~g}$ of $\mathrm{NaOH}$.
Show Answer
Answer: 0.133 MQuestion 19. $0.1 \mathrm{~g}$ of a sample of $\mathrm{KClO} _{3}$ containing some $\mathrm{KCl}$ when decomposed yields oxygen that is sufficient for complete combustion of $40 \mathrm{mLCO}$ at $27^{\circ} \mathrm{C}$ and $750 \mathrm{~mm}$ pressure. Determine the purity of the sample.
Show Answer
Answer: $65 \% $Question 20. A mixture of $\mathrm{HCOOH}$ and $\mathrm{H} _{2} \mathrm{C} _{2} \mathrm{O} _{4}$ is heated with conc. $\mathrm{H} _{2} \mathrm{SO} _{4}$. The gas produced is collected and treated with $\mathrm{KOH}$ solution. The volume of the gas decreased by $1 / 6^{\text {th }}$. Calculate the molar ratio of the two acids in the mixture.
Show Answer
Answer: $1: 4$Question 21. A sample of $\mathrm{NH} _{4} \mathrm{Cl}$ was boiled with $50 \mathrm{~mL}$ of $0.75 \mathrm{~N}$, $\mathrm{NaOH}$ till the reaction was complete. The remaining alkali was neutralized by $10 \mathrm{~mL} 0.75 \mathrm{~N}, \mathrm{H} _{2} \mathrm{SO} _{4}$. What is the amount of $\mathrm{NH} _{4} \mathrm{Cl}$ taken initially?
Show Answer
Answer: $1.605 \mathrm{~g}$Question 22. A gas mixture of $31 \mathrm{~L}$ of propane $\left(\mathrm{C} _{3} \mathrm{H} _{8}\right)$ and butane $\left(\mathrm{C} _{4} \mathrm{H} _{10}\right)$ on complete combustion produced $101 \mathrm{Lof} _{2}$. Find out the composition of the gas mixture.
Show Answer
Answer: $2 \mathrm{~L} \& 1 \mathrm{~L}$Question 23. $3.0 \mathrm{~g}$ of a sample of blue vitriol were dissolved in $\mathrm{H} _{2} \mathrm{O}$. $\mathrm{BaCl} _{2}$ solution was mixed in excess to this solution. The precipitate obtained was washed and dried to $2.8 \mathrm{~g}$. Determine the $\%$ of $\mathrm{SO} _{4}{ }^{2} \cdot$ radical.
Show Answer
Answer: $38.4 $Question 24. The reaction $\mathrm{Zn}+\mathrm{CuSO} _{4} \longrightarrow \mathrm{Cu}+\mathrm{ZnSO} _{4}$ goes completely to the right. In one experiment $10 \mathrm{~g}$ metallic zinc was added to $200 \mathrm{~mL}$ of $\mathrm{CuSO} _{4}$ solution. After all copper is precipitated it was found that not all the Zn had dissolved. After filtration the total solid at the end of the reaction was 9.810 g. Calculate-
i. The weight of copper deposited
ii. Molarity of copper sulphate in the original solution
Show Answer
Answer: (1) $6.55 \mathrm{~g}$, (2) $0.5 \mathrm{M}$Question 25. $1.00 \mathrm{~g}$ of a mixture consisting of equal number of moles of carbonates of the two univalent metals, required $44 \mathrm{~mL}$ of $0.5 \mathrm{~N} \mathrm{HCl}$ for complete reaction. If the atomic weight of one of the metals is 7 , find the atomic weight of the other metal. What is the total amount of sulphate formed, on gravimetric conversion of $1 \mathrm{~g}$ of mixture of carbonate?
