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 m3; Litre (L)
Density kg/m3
Temperature Kelvin (K)
Pressure Pascal (Nm2 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 cm

1 inch 2.54 cm=1=2.54 cm1inch=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

2H2( g)+O2( g)2H2O(g)4g32g36g

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

NO NO2
Mass ratio of N and O 14:16 14:32
7:8 7:16

Ratio of masses of 0 that combines with a fixed mass of N(7 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 (H2O) 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 112 part of an atom of C12.

Relative atomic mass = Mass of one atom of the element  Mass of 1/12 part of atom of C-12 

Average atomic mass =AiXiXtotal 

Where A1, A2, A3 are atomic masses of species 1,2,3.. etc, with their % or fractions X1,X2,X3 etc.

Similar terms are for molar masses.

Calculation of atomic weight

Atomic wt. × specific heat =6.4calK1 mol1 (approx.) (Dulong and Petit’s law for solids)

Approx. atomic weight =6.4/ specific heat

Valency = Atomic mass  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 12C atom

1u=112×12 g mol16.022×1023 mol1

=1.66×1024 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.

Mav=m1xa+m2xb+m3xca+b+c

m1,m2 and m3 are the atomic masses of the isotopic forms of an element in the ratio a,b and c.

Mav is the average atomic mass

If relative abundance is given instead of ratio

Mav=m1xr1+m2xr2+m3xr3r1+r2+r3

r1,r2 and r3 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 (0C or 273 K and one atm or 760 mm of Hg pressure). Its value is 22.4 litre. It is also known as molar volume.

Vapour density :

V.D. = Density of a gas  Density of hydrogen 

= Mass of a certain volume of a gas  Mass of same volume of hydrogen 

2× V.D. = Molecular mass

Solved Examples

Question 1. Convert the 28C temperature into degrees Fahrenheit heat

b) 81.4

a) 77.0

c) 77.4

d) 81.0

Show Answer

Answer : (b)

F=9(C)5+32=95×28+32=81.4

Question 2. What will be the answer of 77.326.3 ?

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 mL, 25.25 mL and 25.0 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 =25.2+25.25+25.03=75.453

=25.15

=25.2 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 g of water

(b) 28 g of CO

(c) 46 g of ethylalcohol

(d) 54 g of nitrogen pentaoxide (N2O5)

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. 2.76 g of silver carbonate on being strongly heating yields a residue weighing

(a) 2.16 g

(b) 2.48 g

(c) 2.32 g

(d) 2.64 g

Show Answer Answer: (a)

6. Which has maximum number of molecules?

(a) 7 g of N2

(b) 2 g of H2

(c) 16 g of NO2

(d) 16 gof 2

Show Answer Answer: (b)

7. 2.6 g of a mixture of calcium carbonate and magnesium carbonate is strongly heated to constant mass 1.3 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) 980mg

(b) 400mg

(c) 1.75 g

(d) 0.74 g

Show Answer Answer: (d)

8. If 0.44 g of a colourless oxide of nitrogen occupies 224 mL at 1520 mmHg and 273C, then the compound is

(a) N2O

(b) NO2

(c) NO4

(d) N2O2

Show Answer Answer: (b)

9. If oxygen is present in one litre flask at pressure of 7.6×1010 mmHg, then the number of oxygen molecules in the flask at 0C will be

(a) 27×1010

(b) 0.27×1010

(c) 0.027×1010

(d) 2.7×1010

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 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 O2 and 1 mole of O respectively.

1 Mole of atoms = Gram atomic mass or molar mass =6.022×1023 molecules =22.4 L at NTP (0C,1 atm)=22.7 L at STP (0C,1 bar )

The number of moles of a substance can be calculated by various means.

Rules in Brief

(1) Number of moles of molecules = Mass ing  Molar mass 

(2) Number of moles of atoms = Mass ing  Molar mass 

(3) Number of moles of gases = Volume at NTP  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

= No. of atoms / molecules/ions/electrons  Avogadro constant 

(5) Number of moles of solute = Molarity x Volume of solution in litres

Percentage Composition

% by weight of solute in solution = mass of solute in gram  mass of solution in gram ×100

% by volume of solute in solution = volume of solute in mL volume of solution in mL×100

% by weight to volume of solute in solution = mass of solute in gram  volume of solution in mL×100

Empirical and Molecular formula:

Steps involved in the calculation of empirical formula:

Convert the mass percentage into grams: considering 100 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,. etc.

n= molar mass  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)

w1= mass of solute,

W2= mass of solvent,

m1= molar mass of solute,

m2= molar mass of solvent, n1= moles of solute

w1+w2= mass of solution,

d= density in gcm3

E1= equivalent mass of solute n2= moles of solvent

(n1+n2)= moles of solution,

V= volume of solution in cm3 or 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 C Mass of solute present in 100 mL of solution Strength C=100w1 V or kgm3
Molarity M Mass of solute present 1 L or 1dm3 of solution Number of mol of solute present in one L(1dm3/ 1000 cm3) of solution M=1000w1 V  g L1 or g dm m13

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 g of solution by mass of solvent =100w1 W1+w2 Dimension-
less
Molality % Mass of solute present in 100 g of solvent Mumber of mol of solute present in one kg solvent M=1000w1w2 m1 mol kg w1
Fraction of a component out of total mol present in a solution X1=n1n1+n2 Dimension- less

Relationship between molality (M’) and Molarity (M)

M=1000M(1000dM1)dM=m11000+1M

Relationship between molality ( M ) and mole fraction (X)

M=1000x1x2 m2

dM=m11000+1M

Relationship between molarity ( M ) and Mole fraction (X)

M=1000w2X1dm2(w1+w2)(1X1)

While diluting a solution from one concentration to another, we can use

M1 V1=M2 V2 (Conentrated solution)  (dilute solution) 

Molarity of mixture :

Let there be three samples of solution (containing same solvent and solute) with their molarity M1,M2,M3 and volumes V1, V2, V3 respectively. These solutions are mixed, molarity of mixed solution may be given as :

M1V1+M2V2+M3V3=M(V1+V2+V3) where, M= resultant molarity 

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

 Eq. wt. of metal = Wt. of metal  Wt. of H2 displaced ×1.008= Wt. of metal  Volume of H2 in mL displaced at NTP ×11200

Oxide formation or reduction of the oxide method:

 Eq. wt. of metal = Wt. of metal  wt. of oxygen combined ×8= wt. of metal  Wt. of O2 displaced/combined in mL at NTP ×5600

Chloride formation method:

 Eq. wt. of metal = Wt. of metal  Wt. of chlorine combined ×35.5= Wt. of metal  Wt. of Cl2 combined in mL at NTP ×11200

Metal displacement method:

 Wt. of metal added to a salt solution  Wt. of metal displaced = Eq. Wt. of metal added  Eq. Wt. of metal displaced 

Double decomposition method:

For a reaction AB+CDAD+BC  Wt. of salt AB added to salt CD (in solution)  Wt. of salt AD precipitated =Eq. Wt. of radical A+Eq. wt. of radical B Eq. Wt. of radical A+Eq. wt. of radical D

Electrolytic method:

Eq. wt=wt deposited by 1 Faraday ( 96500 coulombs)

On passing the same quantity of electricity through two different electrolytic solutions.

 Wt. of X deposited  Wt. of Y deposited = Eq. wt. of X Eq. wt. of Y

Neutralization method:

Eq. w. of an acid = Wt. of the acid neutralized by 1000cc of 1 N base solution

Eq. wt. of a base = Wt. of the base neutralized by 1000cc of 1 N acid solution

For an organic acid (RCOOH)

= Eq. wt. of silver salt (RCOOAg) Eq. Wt. of Silver (108) = Wt. of silver salt  Wt. of silver 

Eq. wt. of acid (RCOOH) = Eq. wt. of RCOOAg - 107

Eq.wt. of an acid = Mol. Wt. of the acid  Basicity 

Eq. wt. of base = Mol. Wt. of the base  Acidity 

Eq.wt. of salt = Mol. Wt. of the salt  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

N= Equivalent of solute  Volume of solution in litre = Weight of solute  Equivalent weight of solute xV in litre 

N=WExV in (L)N=W×1000ExV in mL 

Also, no. of equivalents =NxV( in L)= Wt. of solute  Eq. wt. of solute 

And no. of Milli equivalents =NxV( in mL)= Wt. of solute  Eq. wt. of solute ×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.

aA+bBmM+nN

Meq. of A= Meq. of B= Meq. of M= Meq. of N

(ii) In a compound MxNy

Meq. of MxNy= Meq. of M= Meq. of N

or Eq. of MxNy=Eq. of M=Eq. of N

(iii) In a series of reaction for complete reaction

aA+bBcCdD excess eE excess fFgG

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 H2O2, direct conversions can be made as:

(i) % strength of H2O2=1750× Volume strength of H2O2

(ii) Volume strength of H2O2=5.6× Normality of H2O2

(iii) Volume strength of H2O2=11.2× Molarity of H2O2

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 H&0.

Step III Balance the number of H and 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, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms?

(a) 1.25×102

(b) 2.5×102

(c) 0.02

(d) 3.125×102

Show Answer

Strategy : using the mole concept equate the moles of Mg3(PO4)2 to the number of atoms of oxygen in the formula of the compound.

Solution : 1 mole of Mg3(PO4)2 contains 8 moles of oxygen atoms.

Therefore 0.25 moles of oxygen atoms are present in 1/8×0.25 moles of Mg3(PO4)2 =3.125×102 moles of Mg3(PO4)2

Answer : (d)

Question 2. In the reaction

2Al( aq. )+6HCl( aq. )2Al3+( aq. )+6Cl( aq. )+3H2( g)

(a) 33.6 LHH2( g) is produced regardless of temperature and pressure for every mole of Al that reacts.

(b) 67.2 LH2( g) at NTP (0C,1 atm) is produced for every mole Al that reacts.

(c) 11.2 LH2( g) at NTP (0C,1 atm) is produced for every mole HCl(aq) consumed.

(d) 6 LHCl( aq.) is consumed for every 3LH2( g) produced.

Show Answer

Strategy : use the mole concept where moles of HCl are equated to the volume occupied by H2 gas at N.T.P. as given by the balanced chemical equation.

One mole of a gas at N.T.P. (0C,1 atm) occupies 22.4 L of volume.

Solution : 6 moles of HCl liberate 3 moles of H2 gas.

Convert the moles of H2 gas to the volume occupied by H2 gas =3×22.4 L2 of H2

6 moles of HCl liberate 3×22.4 L of H2

1 mole of HCl will liberate 3×22.46=11.2 L

Answer : (c)

Practice Questions

Question 1. 4.0 g of caustic soda (mol. mass 40 ) contains same number of sodium ions as per present in:

(a) 10.6 g of Na2CO3 (mol. mass 106)

(b) 58.5 g of NaCl (formula mass 58.5)

(c) 100 mL of 0.5MNa2SO4 (formula mass 14)

(d) 1 g equivalent of NaNO3 (mol. mass 85)

Show Answer Answer : (c)

Question 2. 0.1 g of a metal on reaction with dil acid gave 34.2 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 g of a metal on oxidation gave 0.79 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 g of a metal chloride contains 35.5 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 (M) contains 65.5 of chlorine. 100 mL. of the vapour of the chloride of the metal at N.T.P. weights 0.72 g. The molecular formula of the metal chloride is:

(a) MCl

(b) MCl2

(c) MCl3

(d) MCl4

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 M2O3. 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 gMg displaces 1.415 g Cu from the salt solution of Cu. Equivalent weight of 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 Fe2O3 would be :

(a) 18.6

(b) 28

(c) 56

(d) 112.0

Show Answer Answer : (a)

Question 13. 5.3 g of Na2CO3 have been dissolved to make 250 mL of the solution. The normality of the resulting solution will be :

(a) 0.1 N

(b) 0.2 N

(c) 0.4 N

(d) 0.8 N

Show Answer Answer : (c)

Question 14. If 8.3 mL of a sample of H2SO4(36 N) is diluted by 991.7 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 mL of an HCl solution gave 0.1435 g of AgCl when treated with excess of AgNO3. 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 mL of a 0.1 N solution of AgNO3 is added to 500 mL of a 0.1 NKCl solution. The concentration of nitrate in the resulting mixture is :

(a) 0.1 N

(b) 0.05 N

(c) 0.01 N

(d) 0.2 N

Show Answer Answer : (b)

Question 17. The ratio of amounts of H2 S needed to precipitate all the metal ions from 100 mL of 1M AgNO3 and 100 mL of 1MCuSO4 is :

(a) 1:2

(b) 2:1

(c) zero

(d) infinite

Show Answer Answer : (a)

Question 18. 1 L of 18 molar 2SO4 has been diluted to 100 L. The normality of the resulting solution is:

(a) 0.09 N

(b) 0.18 N

(c) 1.80 N

(d) 0.36 N

Show Answer Answer : (d)

Question 19. The normality of 0.3M phosphorus acid (H3PO3) is :

(a) 0.1

(b) 0.9

(c) 0.3

(d) 0.6

Show Answer Answer : (d)

Question 20. One litre of N/2HCl solution was heated in a beaker. When volume was reduced to 600 mL, 3.25 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 Na2SO4HH2O, 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 0.1M2CuSO4 and 0.1MofAl2(SO4)3 is :

(a) 0.2M

(b) 0.7M

(c) 0.8M

(d) 1.2M

Show Answer Answer : (b)

Question 23. The isotopic abundance of C12 and C14 is 98 and 2 respectively. What would be the number of C14 atoms in 12 g carbon sample :

(a) 1.032×1022

(b) 3.01×1023

(c) 5.88×1.023

(d) 6.02×1023

Show Answer Answer : (a)

Question 24. 500 mLof2 contains 6.00×1023 molecules at N.T.P. How many molecules are present in 100 mL of CO2 at N.T.P.?

(a) 6×1023

(b) 1.5×1021

(c) 5.88×1.023

(d) 1.5×1022

Show Answer Answer : (b)

Question 25. What will be the volume of the mixture after the reaction?

NH3( g,1 L)+HCl(g,1.5 L)NH4Cl(s)

(a) 1.5 L

(b) 0.5 L

(c) 1 L

(d) 0 L

Show Answer Answer : (b)

Question 26. What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass =10.8 g ) from the reduction of boron trichloride by hydrogen?

(a) 67.2 L

(b) 44.8 L

(c) 22.4 L

(d) 89.6 L

Show Answer Answer : (a)

Question 27. The number of water molecules present in a drop of water (volume 0.0018 mL ) at room temperature is:

(a) 6.022×1019

(b) 1.084×1018

(c) 4.84×1017

(d) 6.022×1023

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) N2O

(c) CO

(d) CO2

Show Answer Answer : (a)

Question 29. For the formation of 3.65 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 Mg3(PO4)2 will contain 0.25 mole of oxygen atoms?

(a) 1.25×102

(b) 2.5×102

(c) 0.02

(d) 3.125×102

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 g of carbon dioxide

(d) Ten moles of hydrogen

Show Answer Answer : (c)

Question 32. How many grams of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms?

(a) 818.75×103

(b) 818.75×102

(c) 818.75×105

(d) 818.75×103

Show Answer Answer : (b)

Question 33. The molality of 1 L solution of 93H2SO4(w/v) having density 1.84 g/mL is :

(a) 1.043 m

(b) 0.1043 m

(c) 10.43 m

(d) 0.01043 m

Show Answer Answer : (c)

Question 34. 0.7 g of NaCO3xH2O is dissolved in 100 mL water, 20 mL of which required 19.8 mL of 0.1 NHCl. The value of x is :

(a) 4

(b) 3

(c) 2

(d) 1

Show Answer Answer : (c)

Question 35. The normality of 10 mL of a ’ 20 VH2O2 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 g of an oxide of a metal is converted to chloride completely and it yielded 5 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 BaCl2 is mixed with 0.20 mole of Na3PO4, the maximum number of moles of Ba3(PO4)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 g of solvent

(b) 1.0 L of solvent

(c) 1.0 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) 6.022×1023

(b) 1×1023/9.108

(c) 6.022×1023/9.108

(d) 1×108/9.108×6.023

Show Answer Answer : (d)

Question 42. Which has maximum number of atoms? (At. wts are given within brackets)

(a) 24 g of C(12)

(b) 56 g of Fe (56)

(c) 27 g of Al(27)

(d) 108 g of Ag(108)

Show Answer Answer : (a)

Question 43. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is :

(a) 1.78M

(b) 2.00M

(c) 2.05M

(d) 2.22M

Show Answer Answer : (c)

Question 44. The molarity of a solution obtained by mixing 750 mL of 0.5MHCl and 250 mL of 2.0M HCl will be:

(a) 0.875M

(b) 1.00M

(c) 1.75M

(d) 0.0975M

Show Answer Answer : (a)

Question 45. The highest mass corresponds to which of the following :

(a) 1 molecule of O2

(b) 1×1023 g mole of O2

(c) an 02 ion

(d) 1 mole of O2

Show Answer Answer : (d)

Question 46. The chloride of a metal has the formula MCl3. The formula of its phosphate will be:

(a) M2PO4

(b) MPO4

(c) M3PO4

(d) M(PO4)2

Show Answer Answer : (b)

Question 47. By heating 10 gCaCO3,5.6 gCaO is formed, what is the weight of CO2 obtained in this reaction?

(a) 4.4 g

(b) 44 g

(c) 2.2 g

(d) 22 g

Show Answer Answer : (a)

Question 48. If 32 g of O2 contains 6.022×1023 molecules at NTP, then 32 g of S, under the same conditions, will contain :

(a) 6.022×1023 S

(b) 3.011×1023 S

(c) 12.044×1023 S

(d) 1×1023 S

Show Answer Answer : (a)

Question 49. The total number of protons in 10 g of calcium carbonate is

(a) 1.5057×1024

(b) 2.0478×1024

(c) 3.0115×1024

(d) 4.0956×1024

Show Answer Answer : (c)

Question 50. How many millilitres ( mL ) of 1MH2SO4 solution is required to neutralize 10 mL of 1M 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 L or 22400 mL at NTP. The following example shows the mass-volume relationship.

CaCO3+2HCl=CaCl2+H2O+CO2

1 mol 2 mol 1 mol

100 g 73 g 44 g

22.4 L or 22400 mL 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: PV=wMRT

Where wis the mass of the gas, M is the molar mass and 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,

2NO+O2=2NO2

2vol.1vol.2vol. (Gay-Lussac’s law)

Under similar conditions of temperature and pressure, gases react in simple ratio of their volumes.

2NO+O2=2NO22 mol1 mol2 mol2×22.4 litre22.4 litre2×22.4 litre (Avogadro’s law)

Under similar conditions of temperature and pressure, equal moles of gases occupy equal volumes.

Example:

2H2( g) O2( g) 2H2O(g)
Mole ratio 2 1 2
Mass ratio 4 g 32 g 36 g
Volume ratio 2vol 1vol 2vol
Volumeat STP 2×22.4 L 22.4 L 2×22.4 L
Molecular ratio 2×6.022×1023 1×6.022×1023 2×6.022×1023

1. Calculations Based on Mass-Mass Relationship

Calculate the weight of iron which will be converted into its oxide (Fe3O4) by the action of 14.4 g of steam on it.

Solution. The balanced chemical equation is :

3Fe+4H2OFe3O4+4H23 mol4 mol3×564×18=168 g=72 g

Now 72 g of steam react with =168 g of Fe

14.4 g of steam will react with =16872×14.4

=33.6 g of Fe 

2. Based on Mass-Volume Relationship

A sample of lime stone contains 80CaCO3. Calculate the volume of CO2 at NTP obtained by treating 25 g of this sample with excess of dil HCl.

Show Answer

Solution : Mass of pure CaCO3=25×0.8=20 g

CaCO3+2HClCaCl2+H2O+CO2

100 g1mole

22.4LatNTP

100 gCaCO3 gives CO2 at NTP =22.4 L

20 gCaCO3 give CO 2 at NTP =22.4×20100=4.48 L

3. Based on Volume-Volume Relationship

One litre mixture of CO and CO2 is passed over red hot coke. Volume of the gaseous mixture becomes 1.4 litre. Calculate the volume of CO and CO2 in the original mixture. All volumes are taken under identical conditions.

Show Answer

Solution : Suppose Volume of CO2= a litre, Volume of CO=(1a) litre

CO2 reacts with coke (carbon). CO does not react with carbon

CO2+C=2CO

1 litre 2 litre

a litre 2a litre

Volume of CO formed =2 a litres.

Total volume of the gases =(1a)+2a=1.4 litre

a=0.4 litre (volume of CO2 )

1a=10.4=0.6 litre (volume of 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 :

KClO3( s)KCl(s)+O2( 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 Katoms in KClO3= moles of K atoms in KCl.

Now, since 1 molecule of KClO3 contains 1 atom of K or 1 mol of KClO3 contains 1 mole of K and similarly 1 mole of KCl contains 1 mole of K.

Thus, moles of K atoms in KClO3=1× moles of KClO3

And moles of K atoms in KCl=1× moles of KCl

Moles of KClO3= moles of KCl

 Mass of KClO3 in g Molar mass of KClO3= Mass of KCl ing  Molar mass of KCl

The above equation gives the weight relationship between KClO3 and KCl which is important in stoichiometric calculations.

Again applying principle of atoms conservation of 0 atoms,

Moles of O in KClO3= moles of O in O2.

But since 1 mole of KClO3 contains 3 moles of 0 and 1 mole of O2, contains 2 moles of 0 thus moles of O in KClO3=3 x moles of KClO3.

Moles of O in O2=2x moles of O2

Therefore 3x moles of KClO3=2x moles of O2

Or 3× Wt. of KClO3 ing  Mol. Wt. of KClO3=2× Volume of O2 at NTP  Standard Molar Volume 

The above equation thus gives weight-volume relationship of reactant and product.

Question. A sample of KClO3 on decomposition yielded 448 mL of oxygen gas at NTP. Calculate

i. Wt. of oxygen produced

ii. Wt. of KClO3 originally taken and

iii. Wt. of KCl produced

(K=39,Cl=35.5 and 0=16)

Show Answer

Solution :

i. Mole of oxygen =44822400=0.02

Wt. of oxygen =0.02×32=0.64 g

ii. KClO3KCl+O2

Apply POAC for 0 atoms,

Moles of O atoms in KClO3= moles of O atoms in O2

3× moles of KClO3=2x moles of O2

( 1 mole of KClO3 contains 3 moles of 0 and 1 moles of O2 contains 2 moles of 0 )

3× Wt. of KClO3 in g  Mol. Wt. of KClO3=2× Volumeat NTP (litres) 22.4

3× Wt. of KClO3122.5=2×0.44822.4

Wt. of KClO3=1.634 g.

iii. Again applying POAC for K atoms

1× moles of KClO3=1× moles of KCl

( 1 mole of KClO3 contains 1 mole of K and 1 mole of KCl contains 1 mole of K )

1x Wt. of KClO3 Mol. wt. of KClO3=1x Wt. of KCl Mol. wt. of KCl

1.634122.5= Wt. of KCl 74.5

Wt. of KCl =0.9937 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.

A+2 B4CInitially5 mole12 mole0 moleFinally0 mole2 mole20 mole

Hence A is the limiting reactant.

Or The reactant with the least number of equivalents (or milli equivalants) is the limiting reactant.

Question. 500 mL of 0.25MNa2SO4 is added to an aqueous solution of 15 g of BaCl2 resulting in the formation of white precipitate of insoluble BaSO4. How many moles and how many grams of BaSO4 are formed?

Show Answer

Solution : Reaction involved in the process is :

BaCl2+NaSO4BaSO4+2NaCl

1 mol \qquad\qquad 1 mol \qquad\qquad 1 mol \quad \qquad 2mol 

Molarity of Na2SO4 is 0.25 . Thus, 1000 mL solution contains 0.25 mole Na2SO4.

Number of moles of Na2SO4 in 500 mL will be 0.125

Moles of BaCl2= Mass  Molecular mass =15208=0.0721

1 mole Na2SO4 reacts with 1 mole BaCl2

0.125 moles Na2SO4 will react with 0.125 moles BaSO4

Required moles of BaCl2> available moles of BaCl2

Thus BaCl2 is limiting reactant. 1 mole BaCl2 will give 1 mole Na2SO4

0.0721 moles of BaCl2 will give 0.0721 moles of BaSO4

Mass of BaSO4= number of moles x molecular mass

=0.072×233=16.776 g

Neutralization

The term neutralization is used for a reaction between an acid and a base or alkali.

Acid + base = salt + water

For example, HCl+NaOHNaCl+H2O

Neutralization is a quantitative reaction.

A general definition is based on Bronsted & Lowry acid - base theory.

HA+B2+A+BH2+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, OH, the reaction can be written as

HA+OHA+H2O

When the acid has been neutralized there are no molecules of 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, N1 V1=N2 V2

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

N1V1=N2V2

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 810 Weak organic acid
Methyl orange 34.4 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 (HCl) using these two indicators is summarized below:

Phenolphthalein Methyl Orange
NaOH 100 reaction is indicated
NaOH+HClNaCl+H2O
100 reaction is indicated
NaOH+HClNaCl+H2O
Na2CO3 50 reaction upto NaHCO3
Stage is indicated
Na2CO3+HClNaHCO3+NaCl
100 reaction is indicated
Na2CO3+2HCL2NaCl+H2O+CO2
NaHCO3 No reaction is indicated 100 reaction is indicated
NaHCO3+HClNaCl+H2O+CO2

Suppose volume of given standard acid solution (say HCl ) required

for complete reaction of Na2CO3=xmL

for complete reaction of NaHCO3=ymL

for complete reaction of NaOH=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 NaOH+Na2CO3 z+x2 10050 (x+z) 100 each z+x2 10050 x2 (remaining 50% Na2CO3 is indicated)
2 NaOH+NaHCO3 z+0100no reaction (y+z) 100 each (z+0) y { remaining 100NaHCO3 is indicated }
3 Na2CO3+NaHCO3 x2 +0 50 no reaction (x+y) 100 each (x+0)2 (x+y)2 (remaining 50% of Na2CO3 and 100NaHCO3 are indicated)

Solved Examples

Question 1. A solution containing 2.675 g of CoCl3.6NH3 (molar mass =267.5 g/mol ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.76 g of AgCl (molar mass =143.5 g/mol ). The formula of the complex is (At. Mass of Ag= 108u)

(a) {CoCl(NH3)5}Cl2

(b) {Co(NH3)6}Cl3

(c) {CoCl2(NH3)4}Cl

(d) {CoCl3(NH3)3}

Show Answer

Strategy: write a chemical equation of the above process. Calculate the number of moles of AgCl and the number of moles of CoCl3. 6NH3. Using the stoichiometry, equate the moles calculated.

Solution : CoCl36NH3xClAgNO3×AgCl

Number of moles of AgCl=4.78143.6

Number of moles of CoCl3.6NH3=2.675267.5

n(AgCl)=x.n(CoCl3.6NH3)

4.78143.5=x2.675267.5

x=3.333

The complexis [Co(NH3)6]Cl3

Question 2. For a reaction A+2BC, 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 B5 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, 4NH3+5O24NO+6H2O, when one mole of ammonia and one mole of oxygen are made to react to completion, then

(a) 1.0 mole of H2O is produced

(b) all the oxygen is consumed

(c) 1.5 mole of NO is formed

(d) all the ammoniais consumed

Show Answer Answer: (b)

Question 2. 5 mL of 0.1MPb(NO3)2 is mixed with 10 mL of 0.02MKI. The amount of Pbl2 precipitated with be about

(a) 103 mol

(b) 104 mol

(c) 2×104 mol

(d) 4×103 mol

Show Answer Answer: (b)

Question 3. A solution containing Na2CO3 and NaHCO3 is titrated against 0.1MHCl solution using methyl orange indicator. At the equivalence point

(a) Both Na2CO3 and NaHCO3 are completely neutralized

(b) Only Na2CO3 is wholly neutralized

(c) Only NaHCO3 is wholly neutralized

(d) Na2CO3 is neutralized upto the stage of NaHCO3

Show Answer Answer: (a)

Question 4. A solution containing Na2CO3 and NaHCO3 is titrated against 0.1MHCl solution using phenolphthalein indicator. At the equivalence point

(a) Both Na2CO3 and NaHCO3 are completely neutralized

(b) Only Na2CO3 is wholly neutralized

(c) Only NaHCO3 is wholly neutralized

(d) Na2CO3 is neutralized upto the stage of NaHCO3

Show Answer Answer: (d)

Question 5. A solution containing NaOH and Na2CO3 is titrated against 0.1MHCl solution using phenolphthalein indicator. At the equivalence point

(a) Both NaOH and Na2CO3 are completely neutralized

(b) Only NaOH is completely neutralized

(c) Only Na2CO3 is completely neutralized

(d) NaOH completely and Na2CO3 upto the stage of NaHCO3 are neutralized.

Show Answer Answer: (d)

Question 6. A mass of 0.355 g of the compound M2CO3H2O (molar mass of M=23 g mol1 ) is dissolved in 100 mL water and titrated against 0.05MHCl using methyl orange indicator. If the volume of HCl consumed is 100 mL the value of x is

(a) 2

(b) 5

(c) 7

(d) 10

Show Answer Answer: (a)

Question 7. 20 mL of a solution containing 0.34 g of an impure sample of H2O2 reacts with 0.465 g of KMnO4 (molar mass =155 g mol1 ) in acidic medium. The percent purity of H2O2 is about

(a) 60

(b) 70

(c) 75

(d) 80

Show Answer Answer: (c)

Question 8. A mixture of Na2CO3 and NaOH in a solution requires 20 mL of 0.1MHCl solution for neutralization when phenolphthalein indicator is used. The volume consumed is 25 mL when methyl orange indicator is used. The mass per cent of Na2CO3 in the given mixture is about

(a) 47

(b) 57

(c) 67

(d) 75

Show Answer Answer: (a)

Question 9. The mass of AgCl precipitated when 4.68 g of NaCl is added to a solution containing 6.8 g of AgNO3 is

(a) 4.52 g

(b) 5.74 g

(c) 7.18 g

(d) 8.2 g

Show Answer Answer: (b)

Question 10. Twenty milliliter of a solution is 0.1M in each of Na2CO3 and NaHCO3. It is titrated against 0.1M HCl using phenolphthalein as the indicator. The volume of HCl used at the end point will be

(a) 10 mL

(b) 20 mL

(c) 30 mL

(d) 60 mL

Show Answer Answer: (b)

Question 11. An aqueous solution of 6.3 g oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is

(a) 40 mL

(b) 20 mL

(c) 10 mL

(d) 4 mL

Show Answer Answer: (a)

Question 12. Ten milliliter of 0.01M iodine solution is titrated against 0.01M sodium thiosulphate solution using starch solution. The volume of sodium thiosulphate consumed upto the end point is

(a) 10 mL

(b) 15 mL

(c) 20 mL

(d) 30 mL

Show Answer Answer: (c)

Question 13. An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is

(a) 40 mL

(b) 20 mL

(c) 10 mL

(d) 4 mL

Show Answer Answer: (a)

Question 14. In the standardization of Na2 S2O3 using K2Cr2O7 by iodometry, the equivalent mass of K2Cr2O7 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 g oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solutions is :

(a) 40 mL

(b) 20 mL

(c) 10 mL

(d) 4 mL

Show Answer Answer: (a)

Question 16. No. of oxalic acid molecules in 100 mL of 0.02 N oxalic acid are:

(a) 6.022×1020

(b) 6.022×1021

(c) 6.022×1022

(d) 6.022×1023

Show Answer Answer: (a)

Question 17. 30 mL of an acid is neutralized by 15 mL of 0.2 N alkali. The strength of the acid is

(a) 0.1 N

(b) 0.2 N

(c) 0.3 N

(d) 0.4 N

Show Answer Answer: (a)

Question 18. To neutralize 100 mL0.1MH2SO4, the mass of NaOH required is :

(a) 40 g

(b) 80 g

(c) 0.4 g

(d) 0.8 g

Show Answer Answer: (d)

Question 19. 0.16 g of a dibasic acid required 25 mL of decinormal 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 cm3 of 0.5 NH2SO4 to get decinormal concentration is

(a) 400 cm3

(b) 450 cm3

(c) 500 cm3

(d) 100 cm3

Show Answer Answer: (a)

Question 21. The mass of 70H2SO4 required for neutralization of 1 mol of NaOH is

(a) 49 g

(b) 98 g

(c) 70 g

(d) 34 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 kg of it will be

(a) 1.8 kg

(b) 2.7 kg

(c) 4.5 kg

(d) 3.58 kg

Show Answer Answer: (d)

Question 23. 12 L of H2 and 11.2 litres of Cl2 are mixed and exploded. The composition by volume of mixture is

(a) 24 L of HCl

(b) 0.8LCl2 and 20.8 LHCl

(c) 0.8 LH2,22.4 LHCl

(d) 22.4 LHCl

Show Answer Answer: (c)

Question 24. If potassium chlorate is 80 pure, then 48 g of oxygen would be produced from (atomic mass of K=30 )

(a) 153.12 gofKClO3

(b) 122.5 g of KClO3

(c) 245 gofKClO3

(d) 98.0 gof KClO3

Show Answer Answer: (a)

Question 25. The mass of oxygen that would be required to produce enough CO which completely reduces 1.6 kgFe2O3 (at. mass Fe=56 ) is

(a) 240 g

(b) 480 g

(c) 720 g

(d) 960 g

Show Answer Answer: (b)

Question 26. 2.76 g of silver carbonate on being strongly heated yields a residue weighing

(a) 2.16 g

(b) 2.48 g

(c) 2.32 g

(d) 2.64 g

Show Answer Answer: (a)

Question 27. 30 g of Magnesium and 30 g of oxygen are reacted, then the residual mixture contains

(a) 60 g of magnesium oxide only

(b) 40 g of magnesium oxide and 20 g of oxygen

(c) 45 g of magnesium oxide and 15 g of oxygen

(d) 50 g of magnesium oxide and 10 g of oxygen

Show Answer Answer: (d)

Question 28. Minimum volume of SO2 gas at N.T.P. which reduces 100 mL of 0.1NK2Cr2O7 is

(a) 11.2 mL

(b) 22.4 mL

(c) 40 mL

(d) 112 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 g of a sample of pyrolusite (MnO2) when heated with conc. HCl gave chlorine. The chlorine when passed through potassium iodide solution liberated 6.35 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 mL of 0.1 NHCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of the two?

Show Answer Answer: 157.8ml

Question 2. 0.5 g mixture of K2CO3 and Li2CO3 required 30 mL of 0.25 NHCl for neutralization. What is the % composition of mixture. ( K=39,Li=7 )

Show Answer

Answer: % of K2CO3=96

% of Li2CO3=4

Question 3. A solution containing 4.2 g of KOH and Ca(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 KOH=35

% of Ca(OH)2=65

Question 4. A solution contains 4 g of Na2CO3 and NaCl in 250 mL. 25 mL of this solution required 50 mL of N/10HCl for complete neutralization. Calculate % composition of mixture.

Show Answer

Answer: % of Na2CO3=66.25

% of NaCl=33.75

Question 5. What weight of Na2CO3 of 95 purity would be required to neutralize 45.6 mL of 0.235 N acid?

Show Answer Answer: W=0.5978 g

Question 6. Calculate the normality of the resulting solution made by adding 2drops(0.1 mL) of 0.1 NH2SO4 in 1 litre of distilled water.

Show Answer Answer: N=105

Question 7. What weight of AgCl will be precipitated when a solution containing 4.77 gNaCl is added to a solution of 5.77 g of AgNO3 ?

Show Answer Answer: W=4.87 g

Question 8. How much AgCl will be formed by adding 200 mL of 5 NHCl to a solution containing 1.7 g AgNO3 ?

Show Answer Answer: W=1.435 g

Question 9. What is the normality and nature of a mixture obtained by mixing 0.62 g of Na2CO3H2O to 100 mL of 0.1 NH2SO4 ?

Show Answer Answer: 0.1N Neutral

Question 10. 0.5 g of fuming H2SO4 is diluted with water the solution requires 26.7 mL of 0.4 NNaOH for its neutralization. Find the % of freeSO3 in the sample of oleum?

Show Answer Answer: % of SO3=20.78

Question 11. A sample of mixture of CaCl2 and NaCl weighing 4.22 g was heated to precipitate all the Ca2+ ions as CaCO3 which is then quantitatively converted to 0.929 g of CaO. Calculate the % of CaCl2 in the mixture.

Show Answer Answer: 1.84 g,43

Question 12. 10 mL of H2SO4 solution [density 1.84 g/cc ] contains 98H2SO4 by weight. Calculate the volume of 2.5 NNaOH solution required to just neutralize the acid.

Show Answer Answer: 147.2ml

Question 13. One gram of a sample of lime stone was treated with 25 mL of 1 NHCl solution and the volume was made upto 250 mL with water, 25 mL of this solution required 9 mL of 0.1 NNaOH for neutralization. Determine the percentage of CaCO3 in the sample.

Show Answer Answer: 80%

Question 14. If a 20 mL of 0.5 NNaOH solution is mixed with 30 mL of 0.3 NHCl, 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 N

Question 15. 100 mL of a solution of sulphuric acid, 50 mL, N/10 solution of NaOH of density 1.28 g/cc was added and the volume made upto 200 mL,20 mL of this solution required 17 mL of N/10 Na2CO3 solution for neutralization. Determine the normality of sulphuric acid solution.

Show Answer Answer: 0.22 N

Question 16. 1.575 g of oxalic acid (COOH)2×H2O crystal were dissolved in water and the solution made upto 250 mL.20.85 mL of this solution required 25 mL of N/12 NaOH for complete neutralization. Calculate the value of x ?

Show Answer Answer: 2

Question 17. 0.7324 g of zinc dust, containing Zn and ZnO were dissolved in dil H2SO4 liberated 224 mL of H2 at NTP. Calculate the % of Zn metal present in the sample. (Zn=65.4).

Show Answer Answer: 89.3 

Question 18. Find out the molarity of a solution obtained by mixing 2.5 L of 0.1MNaOH,500ML water, 0.05 mole NaOH and 4 g of NaOH.

Show Answer Answer: 0.133 M

Question 19. 0.1 g of a sample of KClO3 containing some KCl when decomposed yields oxygen that is sufficient for complete combustion of 40mLCO at 27C and 750 mm pressure. Determine the purity of the sample.

Show Answer Answer: 65%

Question 20. A mixture of HCOOH and H2C2O4 is heated with conc. H2SO4. The gas produced is collected and treated with KOH solution. The volume of the gas decreased by 1/6th . Calculate the molar ratio of the two acids in the mixture.

Show Answer Answer: 1:4

Question 21. A sample of NH4Cl was boiled with 50 mL of 0.75 N, NaOH till the reaction was complete. The remaining alkali was neutralized by 10 mL0.75 N,H2SO4. What is the amount of NH4Cl taken initially?

Show Answer Answer: 1.605 g

Question 22. A gas mixture of 31 L of propane (C3H8) and butane (C4H10) on complete combustion produced 101Lof2. Find out the composition of the gas mixture.

Show Answer Answer: 2 L&1 L

Question 23. 3.0 g of a sample of blue vitriol were dissolved in H2O. BaCl2 solution was mixed in excess to this solution. The precipitate obtained was washed and dried to 2.8 g. Determine the % of SO42 radical.

Show Answer Answer: 38.4

Question 24. The reaction Zn+CuSO4Cu+ZnSO4 goes completely to the right. In one experiment 10 g metallic zinc was added to 200 mL of CuSO4 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 g, (2) 0.5M

Question 25. 1.00 g of a mixture consisting of equal number of moles of carbonates of the two univalent metals, required 44 mL of 0.5 NHCl 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 g of mixture of carbonate?

Show Answer Answer: 23&1.41


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