Thinking Process

Look at the reading of students $A$ and $B$ given in the question while keeping in mind the concept of precision and accuracy i.e.,

(i) Closeness of reading is precision, and

(ii) If mean of reading is exactly same as the correct value then it is known as accuracy.

Answer

(b) Average of readings of student, $A=\frac{3.01+2.99}{2}=3.00$

Average of readings of student, $B=\frac{3.05+2.95}{2}=3.00$

Correct reading $=3.00$

For both the students, average value is close to the correct value. Hence, readings of both are accurate.

Readings of student $A$ are close to each other (differ only by 0.02 ) and also close to the correct reading, hence, readings of $A$ are precise also. But readings of $B$ are not close to each other (differ by 0.1 ) and hence are not precise.

Answer

(c) There are three common scales to measure temperature ${ }^{\circ} \mathrm{C}$ (degree celsius),

${ }^{\circ} \mathrm{F}$ (degree fahrenheit) and $\mathrm{K}$ (kelvin). The $\mathrm{K}$ is the $\mathrm{SI}$ unit.

The temperatures on two scales are related to each other by the following relationship.

$$ { }^{\circ} \mathrm{F}=\frac{9}{5} t^{\circ} \mathrm{C}+32 $$

Putting the values in above equation

$$ \begin{array}{rlrl} & 200-32 =\frac{9}{5} t^{\circ} \mathrm{C} \\ \Rightarrow & \frac{9}{5} t^{\circ} \mathrm{C} =168 \\ \Rightarrow & t^{\circ} \mathrm{C} =\frac{168 \times 5}{9}=93.3^{\circ} \mathrm{C} \end{array} $$

Answer

(c) Since, molarity $(M)$ is calculated by following equation

$$ \begin{aligned} \text { Molarity } & =\frac{\text { weight } \times 1000}{\text { molecular weight } \times \text { volume }(\mathrm{mL})} \\ & =\frac{5.85 \times 1000}{58.5 \times 500}=0.2 \mathrm{~mol} \mathrm{~L}^{-1} \end{aligned} $$

Note Molarity of solution depends upon temperature because volume of a solution is temperature dependent.

Thinking Process

In case of solution, molarity is calculated by using molarity equation, $M_{1} V_{1}=M_{2} V_{2}$, we have, $V_{1}$ (before dilution) and $V_{2}$ (after dilution), so calculate molarity of the given solution from this equation.

Answer

(b) Given that,

$$ \begin{aligned} M_{1} & =5 \mathrm{M} \\ V_{1} & =500 \mathrm{~mL} \\ V_{2} & =1500 \mathrm{~mL} \\ M_{2} & =M \end{aligned} $$

For dilution, a general formula is

$$ \begin{aligned} M_{1} V_{1} & =M_{2} V_{2} \\ \text { (Before dilution) } & \text { (After dilution) } \\ 500 \times 5 \mathrm{M} & =1500 \times M \\ M= & \frac{5}{3}=1.66 \mathrm{M} \end{aligned} $$

Thinking Process

The number of atoms is related to Avogadro’s number $\left(N_{A}\right)$ by

Number of atoms $=$ moles $\times N_{A}$

The number of atoms of elements can be compared easily on the basis of their moles only because $N_{A}$ is a constant value. Thus, element with large number of moles will possess greatest number of atoms.

Answer

(d) For comparing number of atoms, first we calculate the moles as all are monoatomic and hence, moles $\times N_{A}=$ number of atoms.

$$ \begin{aligned} \text { Moles of } 4 \mathrm{~g} \mathrm{He} & =\frac{4}{4}=1 \mathrm{~mol} \\ 46 \mathrm{~g} \mathrm{Na} & =\frac{46}{23}=2 \mathrm{~mol} \\ 0.40 \mathrm{~g} \mathrm{Ca} & =\frac{0.40}{40}=0.1 \mathrm{~mol} \\ 12 \mathrm{~g} \mathrm{He} & =\frac{12}{4}=3 \mathrm{~mol} \end{aligned} $$

Hence, $12 \mathrm{~g}$ He contains greatest number of atoms as it possesses maximum number of moles.

Answer

(c) In the given question, $0.9 \mathrm{~g} \mathrm{~L}^{-1}$ means that $1000 \mathrm{~mL}$ (or $1 \mathrm{~L}$ ) solution contains $0.9 \mathrm{~g}$ of glucose.

$\therefore$ Number of moles $=0.9 \mathrm{~g}$ glucose $=\frac{0.9}{180}$ mol glucose

$$ =5 \times 10^{-3} \mathrm{~mol} \text { glucose } $$

(where, molecular mass of glucose $\left(C_6 H_12 \mathrm{O}_{6}\right)=12 \times 6+12 \times 1+6 \times 16=180 \mathrm{u}$ ) i.e., $1 \mathrm{~L}$ solution contains 0.05 mole glucose or the molarity of glucose is $0.005 \mathrm{M}$.

Answer

(d) Molality is defined as the number of moles of solute present in $1 \mathrm{~kg}$ of solvent. It is denoted by $m$.

Thus,

$$ \text { Molality }(m)=\frac{\text { Moles of solute }}{\text { Mass of solvent (in } \mathrm{kg})} \quad…(i) $$

Given that, Mass of solvent $\left(\mathrm{H}_{2} \mathrm{O}\right)=500 \mathrm{~g}=0.5 \mathrm{~kg}$

$$ \text { Weight of } \mathrm{HCl}=18.25 \mathrm{~g} $$

Molecular weight of $\mathrm{HCl}=1 \times 1+1 \times 35.5=36.5 \mathrm{~g}$

$$ \begin{aligned} \text { Moles of } \mathrm{HCl} & =\frac{18.25}{36.5}=0.5 \\ m & =\frac{0.5}{0.5}=1 \mathrm{~m} \quad[\text{from Eq. (i)}] \end{aligned} $$

Answer

(a) One mole of any substance contains $6.022 \times 10^{23}$ atoms $/$ molecules.

Hence, Number of millimoles of $ H_2 SO_4 $

$$ \begin{aligned} & =\text { molarity } \times \text { volume in } \mathrm{mL} \\ & =0.02 \times 100=2 \text { millimoles } \\ & =2 \times 10^{-3} \mathrm{~mol} \end{aligned} $$

Number of molecules $=$ number of moles $\times N_{\mathrm{A}}$

$$ \begin{aligned} & =2 \times 10^{-3} \times 6.022 \times 10^{23} \\ & =12.044 \times 10^{20} \text { molecules } \end{aligned} $$

Answer

(b) Molecular mass of $\mathrm{CO}_{2}=1 \times 12+2 \times 16=44 \mathrm{~g}$

$1 \mathrm{~g}$ molecule of $\mathrm{CO}_{2}$ contains $1 \mathrm{~g}$ atoms of carbon

Q $\quad 44 \mathrm{~g}$ of $\mathrm{CO}_{2}$ contain $\mathrm{C}=12 \mathrm{~g}$ atoms of carbon

$\therefore \quad$% of $\mathrm{C}$ in $\mathrm{CO}_{2}=\frac{12}{44} \times 100=27.27 $%

Hence, the mass per cent of carbon in $\mathrm{CO}_{2}$ is $27.27 $%.

Thinking Process

(i) Empirical formula shows that number of moles of different elements present in a molecule, so find the number of moles by dividing molecular mass with empirical formula mass.

(ii) To calculate the molecular formula of the compound, multiply the number of moles with empirical formula.

Answer

(c) Empirical formula mass $=\mathrm{CH}_{2} \mathrm{O}$

$$ \begin{aligned} & =12+2 \times 1+16=30 \\ \text { Molecular mass } & =180 \\ n & =\frac{\text { Molecular mass }}{\text { Empirical formula mass }} \\ & =\frac{180}{30}=6 \end{aligned} $$

$\therefore \quad$ Molecular formula $=n \times$ empirical formula

$$ =6 \times CH_2 O \\ =C_6 H_12 O_6 $$

Answer

(a) Given that, density of solution $=3.12 \mathrm{~g} \mathrm{~mL}^{-1}$

Volume of solution $=1.5 \mathrm{~mL}$

For a solution,

$$ \begin{aligned} \text { Mass } & =\text { volume } \times \text { density } \\ & =1.5 \mathrm{~mL} \times 3.12 \mathrm{~g} \mathrm{~mL}^{-1}=4.68 \mathrm{~g} \end{aligned} $$

The digit 1.5 has only two significant figures, so the answer must also be limited to two significant figures. So, it is rounded off to reduce the number of significant figures.

Hence, the answer is reported as $4.7 \mathrm{~g}$.

Answer

(c) A compound is a pure substance containing two or more than two elements combined together in a fixed proportion by mass and which can be decomposed into its constituent elements by suitable chemical methods.

Further, the properties of a compound are quite different from the properties of constituent elements. e.g., water is a compound containing hydrogen and oxygen combined together in a fixed proportionation. But the properties of water are completely different from its constituents, hydrogen and oxygen.

Thinking Process

This problem is based upon the law of conservation of mass as well as limiting reagent.

(i) Law of conservation of mass is that in which total mass of reactants is equal to total mass of products.

(ii) Limiting reagent represents the reactant which reacts completely in the reaction.

Answer

(a) According to the law of conservation of mass,

Total mass of reactants $=$ Total mass of products

Amount of $Fe_{2} O_{3}$ is decided by limiting reagent.

Answer

(b) In this equation,

$$ \underset{44 ~g}{C_3 H_8(g)}+\underset{32 ~g}{O_2(g)} \longrightarrow \underset{44 ~g}{CO_2(g)}+{\underset{18 ~g}{H_2} O}(g) $$

i.e., mass of reactants $\neq$ mass of products.

Hence, law of conservation of mass is not followed.

Answer

(b) The element, carbon, combines with oxygen to form two compounds, namely, carbon dioxide and carbon monoxide. In $\mathrm{CO}_{2}, 12$ parts by mass of carbon combine with 32 parts by mass of oxygen while in CO, 12 parts by mass of carbon combine with 16 parts by mass of oxygen.

Therefore, the masses of oxygen combine with a fixed mass of carbon (12 parts) in $\mathrm{CO}_{2}$ and $\mathrm{CO}$ are 32 and 16 respectively. These masses of oxygen bear a simple ratio of $32: 16$ or $2: 1$ to each other.

This is an example of law of multiple proportion

Answer

$(a, d)$

1 mole of $ O_{2}$ gas at STP $=6.022 \times 10^{23}$ molecules of $ O_{2}$ (Avogadro number) $=32 ~g^{\text {of } O}_{2}$

Hence, 1 mole of oxygen gas is equal to molecular weight of oxygen as well as Avogadro number.

Answer

( $b, c$ )

For the reaction, $\quad H_{2} SO_{4}+2 NaOH \longrightarrow Na_{2} SO_{4}+2 H_{2} O$

$1 ~L $ of $0.1 M H_{2} SO_{4}$ contains $=0.1$ mole of $H_{2} SO_{4}$

$1 ~L$ of $0.1 M NaOH $ contains $=0.1$ mole of $NaOH $

According to the reaction, 1 mole of $ H_2 SO_4 $ reacts with 2 moles of $NaOH $. Hence, 0.1 mole of $NaOH$ will react with 0.05 mole of $H_{2} SO_{4}$ (and 0.05 mole of $H_{2} SO_{4}$ will be left unreacted), i.e., $NaOH$ is the limiting reactant. Since, 2 moles of $NaOH$ produce 1 mole of $Na_{2} SO_{4}$.

Hence, 0.1 mole of $NaOH$ will produce 0.05 mole of $Na_{2} SO_{4}$.

$\begin{aligned} \text{Mass of} \quad Na_{2} SO_{4} & =\text { moles } \times \text { molar mass } \\ & =0.5 \times(46+32+64) g \\ & =7.10 ~g \\ \text { solution after mixing } & =2 ~L\end{aligned}$

Since, only 0.05 mole of $H_{2} SO_{4}$ is left behind as $NaOH$ completely used in the reaction. Therefore, molarity of the given solution is calculated from moles of $H_{2} SO_{4}$.

$H_{2} SO_{4}$ left unreacted in the solution $=0.05$ mole

$\therefore \quad$ Molarity of the solution $=\frac{0.05}{2}=0.025 ~mol ~L^{-1}$

Answer

$(c, d)$

(c) Number of atoms in $28 \mathrm{~g}$ of $N_{2}=\frac{28}{28} \times N_{A} \times 2=2 N_{A}$ (where, $N_{A}=$ Avogadro number)

Number of atoms in $32 g$ of $O_2 = \frac{32}{32} \times N_A \times 2 = 2N_A $

(d) $12 \mathrm{~g}$ of $\mathrm{C}(\mathrm{s})$ contains atoms $=\frac{12}{12} \times N_{\mathrm{A}} \times 1=\mathrm{N}_{\mathrm{A}}$

Number of atoms in $23 \mathrm{~g}$ of $\mathrm{Na}(s)=\frac{23}{23} \times N_{A} \times 1=N_{A}$

Answer

$(a, b)$

(a) Molarity $(M)=\frac{\text { weight of } \mathrm{NaOH} \times 1000}{\text { Molecular weight of } \mathrm{NaOH} \times V(\mathrm{~mL})}$

$$ =\frac{20 \times 1000}{40 \times 200}=2.5 \mathrm{M} $$

(b) $M=\frac{0.5 \times 1000}{200}=2.5 \mathrm{M}$

(c) $M=\frac{40 \times 1000}{10 \times 100}=10 \mathrm{M}$

(d) $M=\frac{20 \times 1000}{56 \times 200}=1.785 \mathrm{M}$

Thus, $20 \mathrm{~g} \mathrm{NaOH}$ in $200 \mathrm{~mL}$ of solution and $0.5 \mathrm{~mol}$ of $\mathrm{KCl}$ in $200 \mathrm{~mL}$ have the same concentration.

Answer

(c, d)

The number of molecules can be calculated as follows

Number of molecules $=\frac{\text { Mass }}{\text { Molar mass }} \times$ Avogadro number $\left(N_{A}\right)$

Number of molecules, in $16 ~g$ oxygen $=\frac{16}{32} \times N_{A}=\frac{N_{A}}{2}$

In $16 ~g$ of $CO=\frac{16}{28} \times N_{A}=\frac{N_{A}}{1.75}$

In $28 ~g$ of $N_{2}=\frac{28}{28} \times N_{A}=N_{A}$

In $14 ~g$ of $N_{2}=\frac{14}{28} \times N_{A}=\frac{N_{A}}{2}$

In $1 ~g$ of $H_{2}=\frac{1}{2} \times N_{A}=\frac{N_{A}}{2}$

So, $16 ~g$ of $O_{2}=14 ~g$ of $N_{2}=1.0 ~g$ of $H_{2}$

Answer

$(c, d)$

Both mole fraction and mass per cent are unitless as both are ratios of moles and mass respectively.

$$ \begin{aligned} \text { Mole fraction } & =\frac{\text { Number of moles of solute }}{\text { Number of moles of solution }}=\frac{\text { moles }}{\text { moles }} \\ & =\frac{\text { Number of moles of solvent }}{\text { Number of moles of solution }}=\frac{\text { moles }}{\text { moles }} \\ \text { Mass per cent } & =\frac{\text { Mass of solute in gram }}{\text { Mass of solution in gram }} \times 100 \end{aligned} $$

Answer

$(a, d)$

Law of conservation of mass is simply the law of indestructibility of matter during physical or chemical changes. Avogadro law states that equal volumes of different gases contain the same number of molecules under similar conditions of temperature and pressure.

Answer

The mass of a carbon-12 atom was determined by a mass spectrometer and found to be equal to $1.992648 \times 10^{-23} \mathrm{~g}$. It is known that 1 mole of $\mathrm{C}-12$ atom weighing $12 \mathrm{~g}$ contains $\mathrm{N}_{\mathrm{A}}$ number of atoms. Thus,

1 mole of $\mathrm{C}-12$ atoms $=12 \mathrm{~g}=6.022 \times 10^{23}$ atoms

$\Rightarrow \quad 6.022 \times 10^{23}$ atoms of $\mathrm{C}-12$ have mass $=12 \mathrm{~g}$

$\therefore \quad 1$ atom of $\mathrm{C}-12$ will have mass $=\frac{12}{6.022 \times 10^{23}} \mathrm{~g}$

$$ =1.992648 \times 10^{-23} \mathrm{~g} \approx 1.99 \times 10^{-23} \mathrm{~g} $$

Thinking Process

(i) To answer the given calculations, least precise term decide the significant figures.

(ii) To round up a number, left the last digit as such, if the digit next to it is less than 5 and increase it by 1 , if the next digit is greater than 5.

Answer

Least precise term 2.5 or 3.5 has two significant figures.

Hence, the answer should have two significant figures

$$ \frac{2.5 \times 1.25 \times 3.5}{2.01} \approx 5.4415=5.4 $$

Answer

Symbol for SI unit of mole is mol.

One mole is defined as the amount of a substance that contains as many particles and there are atoms in exactly $12 \mathrm{~g}(0.012 \mathrm{~kg})$ of the ${ }^{12} \mathrm{C}$ - isotope.

$$ \frac{1}{12} \mathrm{~g} \text { of }{ }^{12} \mathrm{C} \text {-isotope }=1 \text { mole } $$

Answer

Molality It is defined as the number of moles of solute dissolved in $1 \mathrm{~kg}$ of solvent. It is independent of temperature.

Molarity It is defined as the number of moles of solute dissolved in $1 \mathrm{~L}$ of solution. It depends upon temperature (because, volume of solution $\propto$ temperature).

Thinking Process

To calculate the mass per cent of atom, using the formula

Mass per cent of an element

$$ =\frac{\text { Atomic mass of the element present in the compound }}{\text { Molar mass of the compound }} \times 100 $$

Answer

Mass per cent of calcium $\frac{3 \times \text{(atomic mass of calcium)}}{\text{molecular mass of} Ca_3(PO_4)_2} \times 100$

$ =\frac{120 u}{310 u} \times 100=38.71 $%

$ \text { Mass per cent of phosphorus }=\frac{2 \times(\text { atomic mass of phosphorus })}{\text { molecular mass of } Ca_3 \left( PO_{4}\right)_{2}} \times 100 $

$ =\frac{2 \times 31 u}{310 u} \times 100=20 $%

$ \text { Mass per cent of oxygen }=\frac{8 \times(\text { atomic mass of oxygen })}{\text { molecular mass of } Ca_{3}\left(PO_{4}\right)_{2}} \times 100 $

$ =\frac{8 \times 16 u}{310 u} \times 100=41.29 $%

Answer

For the reaction,

$\underset{2 ~V}2N_{2}(g)+ \underset{1 ~V}{O_{2}(g)} \longrightarrow \underset{2 ~V}2N_2 O(g) $

$ \frac{45.4}{22.7}=2 \quad \frac{22.7}{22.7}=1 \quad \frac{45.4}{22.7}=2 $

Hence, the ratio between the volumes of the reactants and the product in the given question is simple i.e., $2: 1: 2$. It proves the Gay-Lussac’s law of gaseous volumes.

Note Gay-Lussac’s law of gaseous volumes, when gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume provided all gases are at same temperature and pressure.

Answer

(a) Yes, the given statement is true.

(b) According to the law of multiple proportions

(c) $\underset{2 g}{H_{2}}+\underset{16 {~g}}{O_{2}} \longrightarrow \underset{18 {~g}}{H_{2} {O}}$

$\underset{2 g}{H_{2}}+\underset{32 g}{O_{2}} \longrightarrow \underset{34 g}{H_{2} O_{2}}$

Here, masses of oxygen, (i.e., $16 ~g$ in $H_{2} {O}$ and $32 {~g} in_{2} O_{2}$ ) which combine with fixed mass of hydrogen $(2 {~g})$ are in the simple ratio $i . e ., 16: 32$ or $1: 2$.

Answer

Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (per cent occurrence), the average atomic mass of the element can be calculated as

$\lbrace ( \text{Natural abundance of} ^1H \times molar mass)+ \rbrace $

$\text{ Average atomic mass } =\frac{(\text{ Natural abundance of }^{2} {H} \times \text{ molar mass of }^{2} {H})}{100} $

$=\frac{99.985 \times 1+0.015 \times 2}{100} $

$ =\frac{99.985+0.030}{100}=\frac{100.015}{100}=1.00015 {u}$

Answer

Given that, Mass of $\mathrm{Zn}=32.65 \mathrm{~g}$

1 mole of gas occupies $=22.7 \mathrm{~L}$ volume at STP

Atomic mass of $\quad \mathrm{Zn}=65.3 \mathrm{u}$

The given equation is

$$ \underset{65.3 ~g}{Zn}+2 {HCl} \longrightarrow ZnCl_{2}+\underset{1 ~mol=22.7 {~L} \text{ at STP }}{H_{2}} $$

From the above equation, it is clear that $65.3 \mathrm{~g} \mathrm{Zn}$, when reacts with $\mathrm{HCl}$, produces $=22.7 \mathrm{~L}$ of $\mathrm{H}_{2}$ at STP

$\therefore 32.65 \mathrm{~g} \mathrm{Zn}$, when reacts with $\mathrm{HCl}$, will produce $=\frac{22.7 \times 32.65}{65.3}=11.35 \mathrm{~L}$ of $\mathrm{H}_{2}$ at STP.

Thinking Process

Determine the mass of solution from the given molality of the solution followed by volume of solution relating mass and density to each other, i.e.,

$$ \text { Volume }=\frac{\text { Mass }}{\text { Density }} $$

Then, calculate the molarity of solution as

$$ \text { Molarity }=\frac{\text { Number of moles }}{\text { Volume in litres }} $$

Answer

3 molal solution of $\mathrm{NaOH}$ means 3 moles of $\mathrm{NaOH}$ are dissolved in $1 \mathrm{~kg}$ solvent. So, the mass of solution $=1000 \mathrm{~g}$ solvent $+120 \mathrm{~g} \mathrm{NaOH}=1120 \mathrm{~g}$ solution

(Molar mass of $\mathrm{NaOH}=23+16+1=40 \mathrm{~g}$ and 3 moles of $\mathrm{NaOH}=3 \times 40=120 \mathrm{~g}$ )

$$ \begin{aligned} & \text { Volume of solution }=\frac{\text { Mass of solution }}{\text { Density of solution }} \quad \left(\mathrm{Q} d=\frac{m}{V}\right)\\ & \qquad \begin{aligned} V & =\frac{1120 \mathrm{~g}}{1.110 \mathrm{~g} \mathrm{~mL}^{-1}}=1009 \mathrm{~mL} \\ \text { Molarity } & =\frac{\text { Moles of solute } \times 1000}{\text { Volume of solution }(\mathrm{mL})} \\ & =\frac{3 \times 1000}{1009}=2.973 \mathrm{M} \approx 3 \mathrm{M} \end{aligned} \end{aligned} $$

Answer

No, molality of solution does not change with temperature since mass remains unaffected with temperature.

$$ \text { Molality, } m=\frac{\text { moles of solute }}{\text { weight of solvent (in g) }} \times 1000 $$

Thinking Process

(i) To proceed the calculation, first calculate the number of moles of $\mathrm{NaOH}$ and $\mathrm{H}_{2} \mathrm{O}$

(ii) Then, find mole fraction of $\mathrm{NaOH}$ and $\mathrm{H}_{2} \mathrm{O}$ by using the formula,

$$ X_{\mathrm{NaOH}}=\frac{n_{\mathrm{NaOH}}}{n_{\mathrm{NaOH}}+n_{\mathrm{H}_{2} \mathrm{O}}} $$

$$ \left(\operatorname{or} X_{H_{2} {O}}=\frac{n_{H_{2} {O}}}{n_{{NaOH}}+n_{H_{2} {O}}}\right) $$

(iii) Then, calculate molarity $=\frac{W \times 1000}{m \times V}$, so in order to calculate molarity we require volume of solution which is, $V=\frac{m}{\text { specific gravity }}$.

Answer

Number of moles of $\mathrm{NaOH}$,

$n_{NaOH}=\frac{4}{40}=0.1 {~mol} \quad \quad$ $ \lbrace Qn = \frac{Mass(g)}{\text{Molar Mass} (g mol^{-1})} \rbrace$

$Similarly, \quad n_{NaOH}=\frac{36}{18}=2 {~mol} $

$\text { Mole fraction of } {NaOH}, X_{{NaOH}}=\frac{\text { moles of } {NaOH}}{\text { moles of } {NaOH}+\text { moles of } H_{2} {O}} $

$X_{NaOH}=\frac{0.1}{0.1+2}=0.0476 $

$\text{Similarly,} X_{H_{2} {O}}=\frac{n_{H_{2} {O}}}{n_{NaOH}+n_{H_{2} {O}}} $

$=\frac{2}{0.1+2}=0.9524 $

Total mass of solution $=$ mass of solute + mass of solvent

$$ \begin{aligned} & =4+36=40 \mathrm{~g} \\ \text { Volume of solution } & =\frac{\text { Mass of solution }}{\text { specific gravity }}=\frac{40 \mathrm{~g}}{1 \mathrm{~g} \mathrm{~mL}^{-1}}=40 \mathrm{~mL} \\ \text { Molarity } & =\frac{\text { Moles of solute } \times 1000}{\text { Volume of solution }(\mathrm{mL})} \\ & =\frac{0.1 \times 1000}{40}=2.5 \mathrm{M} \end{aligned} $$

Answer

$$ 2 A+4 B \longrightarrow 3 C+4 D $$

According to the given reaction, 2 moles of $A$ react with 4 moles of $B$.

Hence, 5 moles of $A$ will react with 10 moles of $B\left(\frac{5 \times 4}{2}=10\right.$ moles $)$

(a) It indicates that reactant $B$ is limiting reagent as it will consume first in the reaction because we have only 6 moles of $B$.

(b) Limiting reagent decide the amount of product produced.

According to the reaction,

4 moles of $B$ produces 3 moles of $C$

$\therefore 6$ moles of $B$ will produce $\frac{3 \times 6}{4}=4.5$ moles of $C$.

Note Limiting reagent limits the amount of product formed because it is present in lesser amount and gets consumed first.

Answer

A. $\rightarrow(2)$

B. $\rightarrow$ (3)

C. $\rightarrow(1)$

D. $\rightarrow$ (5)

E. $\rightarrow(4)$

A. Number of moles of $CO_{2}$ molecule $=\frac{\text { Weight in gram of } CO_{2}}{\text { Molecular weight of } CO_{2}}=\frac{88}{44}=2 {~mol}$

B. 1 mole of a substance $=N_{A}$ molecules $=6.022 \times 10^{23}$ molecules

$=$ Avogadro number

$=6.022 \times 10^{23}$ molecules of $H_{2} {O}=1 {~mol}$

C. $22.4 {~L}$ of $O_{2}$ at ${STP}=1 {~mol}$

$5.6 \mathrm{~L}$ of $\mathrm{O}_{2}$ at $\mathrm{STP}=\frac{5.6}{22.4} \mathrm{~mol}=0.25 \mathrm{~mol}$

D. Number of moles of $96 \mathrm{~g}^{\text {of }} \mathrm{O}_{2}=\frac{96}{32} \mathrm{~mol}=3 \mathrm{~mol}$

E. 1 mole of any gas = Avogadro number $=6.022 \times 10^{23}$ molecules

Answer

A. $\rightarrow$ (5)

B. $\rightarrow$ (4)

C. $\rightarrow(2)$

D. $\rightarrow(7)$

E. $\rightarrow(3)$

F. $\rightarrow(6)$

G. $\rightarrow(1)$

H. $\rightarrow(9)$

A. Molarity $=$ concentration in $\mathrm{mol} \mathrm{L}^{-1}$

$$ \text { Molarity }=\frac{\text { Number of moles }}{\text { Volume in litres }} $$

B. Mole fraction = Unitless

C. Mole $=\frac{\text { Mass }(\mathrm{g})}{\text { Molar mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)}=\mathrm{mol}$

D. Molality $=$ concentration in mol per $\mathrm{kg}$ solvent

Molality $=\frac{\text { Number of moles }}{\text { Mass of solvent }(\mathrm{kq})}$

$\mathrm{E}$. The $\mathrm{SI}$ unit for pressure is the pascal $(\mathrm{Pa})$, equal to one newton per square metre $\left(\mathrm{N} / \mathrm{m}^{2}\right.$ or $\left.\mathrm{kg} \cdot \mathrm{m}^{-1} \mathrm{~s}^{-2}\right)$. This special name for the unit was added in 1971; before that, pressure in $\mathrm{SI}$ was expressed simply as $\mathrm{N} / \mathrm{m}^{2}$.

F. Unit of luminous intensity = candela.

The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency $540 \times 10^{12}$ hertz and that has a radiant intensity in that direction of $1 / 683$ watt per steradian.

G. Density $=\frac{\text { mass }}{\text { volume }}=\mathrm{g} \mathrm{mL}^{-1}$

$H$. Unit of mass = kilogram

The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram

Assertion and Reason

In the following questions a statement of Assertion (A) followed by a statement of Reason ( $\mathrm{R}$ ) is given. Choose the correct option out of the choices given below in each question.

Answer

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

The molecular formula of ethene is $C_{2} H_{4}$ and its empirical formula is $CH_{2}$.

Thus, Molecular formula $=$ Empirical formula $\times 2$

Answer

(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Atomic masses of the elements obtained by scientists by comparing with the mass of carbon comes out to be close to whole number value.

Answer

(c) Assertion is true but Reason is false.

0.200 contains 3 while 200 contains only one significant figure because zero at the end or right of a number are significant provided they are on the right side of the decimal point.

Answer

(c) Assertion is false but Reason is true.

Combustion of $16 \mathrm{~g}$ of methane gives $36 \mathrm{~g}$ of water.

$\underset{=16g}{\underset{ 1 mol }{CH_{4}}}+{2O_{2}} \longrightarrow CO_2 + \underset{=36g}{\underset{2 {mol}}{2H_{2} {O}}}$

Answer

(a) $p_{1}=1 \mathrm{~atm}, p_{2}=\frac{1}{2}=0.5 \mathrm{~atm}, T_{1}=273.15, V_{2}=$ ? , $V_{1}=$ ?

$32 \mathrm{~g}$ dioxygen occupies $=22.4 \mathrm{~L}$ volume at STP

$\therefore \quad 1.6 \mathrm{~g}$ dioxygen will occupy $=\frac{22.4 \mathrm{~L} \times 1.6 \mathrm{~g}}{32 \mathrm{~g}}=1.12 \mathrm{~L}$

$$ V_{1}=1.12 \mathrm{~L} $$

From Boyle’s law (as temperature is constant),

$$ \begin{aligned} p_{1} V_{1} & =p_{2} V_{2} \\ V_{2} & =\frac{p_{1} V_{1}}{p_{2}} \end{aligned} $$

$$ =\frac{1 \mathrm{~atm} \times 1.12 \mathrm{~L}}{0.5 \mathrm{~atm}}=2.24 \mathrm{~L} $$

(b) Number of moles of dioxygen $=\frac{\text { Mass of dioxygen }}{\text { Molar mass of dioxygen }}$

$$ \begin{aligned} n_{O_{2}} & =\frac{1.6}{32}=0.05 {~mol} \\ 1 {~mol} \text { of dioxygen contains } & =6.022 \times 10^{23} \text { molecules of dioxygen } \\ \therefore \quad 0.05 \text { mol of dioxygen } & =6.022 \times 10^{23} \times 0.05 \text { molecule of } O_{2} \\ & =0.3011 \times 10^{23} \text { molecules } \\ & =3.011 \times 10^{22} \text { molecules } \end{aligned} $$

Answer

Molar mass of $\mathrm{CaCO}_{3}=40+12+3 \times 16=100 \mathrm{~g} \mathrm{~mol}^{-1}$

Moles of $CaCO_{3}$ in $1000 {~g}, n_{CaCO_{3}}=\frac{\text { Mass }({g})}{\text { Molar mass }}$

$$ \begin{aligned} n_{\mathrm{CaCO}_{3}} & =\frac{1000 \mathrm{~g}}{100 \mathrm{~g} \mathrm{~mol}^{-1}}=10 \mathrm{~mol} \\ \text { Molarity } & =\frac{\text { Moles of solute }(\mathrm{HCl}) \times 1000}{\text { Volume of solution }} \end{aligned} $$

(It is given that moles of $\mathrm{HCl}$ in $250 \mathrm{~mL}$ of $0.76 \mathrm{M} \mathrm{HCl}=n_{\mathrm{HCl}}$ )

$$ 0.76=\frac{n_{\mathrm{HCl}} \times 1000}{250} $$

$$ \begin{aligned} & n_{HCl}=\frac{0.76 \times 250}{1000}=0.19 {~mol} . \\ & \underset{1 {~mol}^{CaCO_{3}}({~s})}{CHCl_{2} {Hol}({aq})} \longrightarrow CaCl_{2}({aq})+CO_{2}(g)+H_{2} {O}(l) \end{aligned} $$

According to the equation,

1 mole of $\mathrm{CaCO}_{3}$ reacts with 2 moles $\mathrm{HCl}$

$\therefore \quad 10$ moles of $\mathrm{CaCO}_{3}$ will react with $\frac{10 \times 2}{1}=20$ moles $\mathrm{HCl}$.

But we have only 0.19 moles $\mathrm{HCl}, \mathrm{so} \mathrm{HCl}$ is limiting reagent and it limits the yield of $\mathrm{CaCl}_{2}$.

Since,

2 moles of $\mathrm{HCl}$ produces 1 mole of $\mathrm{CaCl}_{2}$

0.19 mole of $\mathrm{HCl}$ will produce $\frac{1 \times 0.19}{2}=0.095 \mathrm{~mol} \mathrm{CaCl}_{2}$

Molar mass of $\mathrm{CaCl}_{2}=40+(2 \times 35.5)=111 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore \quad 0.095$ mole of $\mathrm{CaCl}_{2}=0.095 \times 111=10.54 \mathrm{~g}$

Answer

‘Law of multiple proportions’ was first studied by Dalton in 1803 which may be defined as follows

When two elements combine to form two or more chemical compounds, then the masses of one of the elements which combine with a fixed mass of the other, bear a simple ratio to one another.

e.g., hydrogen combines with oxygen to form two compounds, namely, water and hydrogen peroxide.

$\underset{2g}{Hydrogen} + \underset{16g}{Oxygen} \rightarrow \underset{18g}{Water}$

$\underset{2g}{Hydrogen} + \underset{32g}{Oxygen} \rightarrow \underset{34g}{\text{Hydrogen peroxide}}$

Here, the masses of oxygen (i.e., $16 \mathrm{~g}$ and $32 \mathrm{~g}$ ) which combine with a fixed mass of hydrogen $(2 \mathrm{~g})$ bear a simple ratio, i.e., $16: 32$ or $1: 2$.

As we know that, when compounds mixed in different proportionation, Then they form different compounds. In the above examples, when hydrogen is mixed with different proportion of oxygen, then they form water or hydrogen peroxide.

It shows that there are constituents which combine in a definite proportion. These constituents may be atoms. Thus, the law of multiple proportions shows the existence of atoms which combine into molecules.

Thinking Process

In this question, it is seen that the masses of $B$ which combine with the fixed mass of $A$ in different combinations are related to each other by simple whole numbers.

Answer

Mass of $B$ which is combined with fixed mass of $A$ (say $1 \mathrm{~g}$ ) will be $2.5 \mathrm{~g}, 5 \mathrm{~g}, 1.25 \mathrm{~g}$ and $3.75 \mathrm{~g}$. They are in the ratio $2: 4: 1: 3$ which is a simple whole number ratio. Hence, the law of multiple proportions is applicable.