Mole Concept Previous Year Questions with Solutions are given here. When dealing with particles at an atomic (or molecular) level, even one gram of a pure element will contain a huge number of atoms. This is where the mole concept is widely used. It focuses on the unit known as a ‘mole’, which is a count of a very large number of particles. A mole is defined as the amount of a substance that contains exactly 6.02214076 × 10^{23} ‘elementary entities’ of the given substance. The number 6.02214076× 10^{23} is known as the Avogadro constant and is denoted by the *symbol ‘NA’*.

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The word “mole” was introduced around the year 1896 by the German Chemist Wilhelm Ostwald, who derived the term from the Latin word moles meaning a ‘heap’ or ‘pile. The mole concept is a convenient method of expressing the amount of a substance. Any measurement can be broken down into two parts – the numerical magnitude and the units that the magnitude is expressed in. For example, when the mass of a sphere is measured to be 2 kilograms, the magnitude is ‘2’ and the unit is ‘kilogram’.

The chapter contains important topics such as atomic and molecular mass, molar mass, gram atomic mass and gram molecular mass. BYJU’S provides accurate solutions prepared by our specialised experts.

## JEE Main Mole Concept Previous Year Questions With Solutions

**1. Number of atoms in the following samples of substances is the largest in :**

(1) 127.0g of iodine

(2) 48.0g of magnesium

(3) 71.0g of chlorine

(4) 4.0g of hydrogen

**Solution:**

1 mole represents 6.023×10^{23} particles.

1 mole of iodine atom= 6.023×10^{23}

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×10^{23}no.of Mg

Given 48g of Mg = 2×6.023×10^{23 }

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 10^{23}

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 10^{23}

no. of chlorine atom = 6.023×10^{23}

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 10^{23}

no. of atoms of hydrogen= 4 moles of hydrogen atom.

Hence option(4) is the answer.

**2. The ratio of mass percent of C and H of an organic compound (C _{x}H_{y}O_{z}) is 6 : 1. If one molecule of the above compound ( C_{x}H_{y}O_{Z }) contains half as much oxygen as required to burn one molecule of compound C_{X}H_{Y} completely to CO_{2} and H_{2}O. The empirical formula of compound C_{x}H_{y}O_{z} is :**

(1) C_{2}H_{4}O

(2) C_{3}H_{4}O_{2}

(3) C_{2}H_{4}O_{3}

(4) C_{3}H_{6}O_{3}

**Solution:**

Given the ratio of mass percent of C and H of an organic compound (C_{x}H_{y}O_{z}) is 6 : 1.

Atomic mass of carbon = 12

Atomic mass of Hydrogen = 1

If we have x atoms of Carbon and y atoms of Hydrogen,

12*x = 6(1*y)

12x = 6y

So y = 2x

Given one molecule of compound ( C_{x}H_{y}O_{Z }) contains half as much oxygen as required to burn one molecule of compound C_{X}H_{Y} completely to CO_{2} and H_{2}O

C_{X}H_{y} + O_{2} → xCO_{2}+ (y/2)H_{2}O

Put y = 2x in above equation

C_{X}H_{2x} + O_{2} → xCO_{2}+ xH_{2}O

Oxygen needed = 2x+x = 3x

z is half of oxygen required to burn.

So z = 3x/2 = 1.5 x

Check the given options which satisfies z = 1.5x.

So the empirical formula is C_{2}H_{4}O_{3}.

Hence option (3) is the answer.

**3. The concentrated sulphuric acid that is peddled commercially is 95% H _{2}SO_{4} by weight. If the density of this commercial acid is 1.834 g cm^{-3}, the molarity of this solution is :-**

(1) 17.8 M

(2) 15.7 M

(3) 10.5 M

(4) 12.0 M

**Solution:**

Given Density = 1.834

1 ml solution contains 1.834 g

1000 ml solution will contain 1834 g

95% H_{2}SO_{4} means 100 gm contain 95 gm H_{2}SO_{4}

Mass of solute = (95/100)×1834

Molecular weight of H_{2}SO_{4} = 98

Molarity = No. of moles/ volume = mass of solute/98

= (95/100)×(1834/98)

= 17.8 M

Hence option (1) is the answer.

**4. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is :**

(1) 1 : 8

(2) 3 : 16

(3) 1 : 4

(4) 7 : 32

**Solution:**

Given ratio of masses of oxygen and nitrogen = 1:4

Let mass of O_{2} = w

Mass of N_{2} = 4w

Molecules of O_{2} = w/(32×N_{A})

Molecules of N_{2} = 4w/(28×N_{A})

Ratio of number of molecules = w/(32×N_{A})÷4w/(28×N_{A})

= w/(32×N_{A})×(28×N_{A})/4w

= 7/32

So the ratio is 7:32.

Hence option (4) is the answer.

**5. 3g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :**

(1) 42 mg

(2) 54 mg

(3) 18 mg

(4) 36 mg

**Solution:**

Molarity of CH_{3}COOH solution = mass of acetic acid/molar mass)/volume of solution in litre

Acetic acid is monobasic.

0.042 = W/(60×0.05)

W = 0.042×60×0.05 = 0.126 g

Amount of acetic acid actually adsorbed = 0.180-0.126 = 54mg

Amount of charcoal available = 3 g

So amount of acetic acid adsorbed per gram of charcoal = 54mg×1g/3.0g = 18 mg

Hence option (3) is the answer.

**6. The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is**

(1) 2.05 M

(2) 0.50 M

(3) 1.78 M

(4) 1.02 M

**Solution:**

Given density of solution = 1.15g/mL

mass of solution = 1000+120 = 1120 gm

Molar mass = 60

Volume = mass /density of solution

= 1120/1.15

No. of moles = 120/60 = 2

Molarity = No. of moles/ volume

= 2÷ (1120×10^{-3}/1.15)

= 2×1.15×1000 /1120)

= 2.05 M

Hence option (1) is the answer.

**7. The ratio of number of oxygen atoms (O) in 16.0g oxygen (O), 28.0 g carbon monoxide (CO) and 32.0g oxygen (O _{2}) is :**

(Atomic mass :C =12, O =16 and Avogadro’s constant N_{A} = 6.0 * 10^{23} mol^{-1})

(1) 3 : 1 : 1

(2) 1 : 1 : 2

(3) 3 : 1 : 2

(4) 1 : 1 : 1

**Solution:**

Molar mass of O_{3} = 48

Given 16 g O_{3} . So no. of moles of O_{3} = 16/48 = ⅓

1 mole = 3 ×N_{A} oxygen atoms

So 1/3 mole = N_{A}×3×1/3 no of atoms

= N_{A} oxygen atoms

Molar mass of CO = 28

Given 28 g CO. So no of moles = 28/28 = 1

No. of atoms = 1×N_{A} = N_{A }

Molar mass of O_{2} = 32

Given 32g O_{2}

No. of moles = 32/32 = 1

No.of atoms = 1×N_{A} = N_{A }

So the ratio is 1:1:1

Hence option (4) is the answer.

**8. When CO _{2} (g) is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5 litre of CO_{2} (g) over red hot coke, the total volume of the gases increased to 700 mL. The composition of the gaseous mixture at STP is :**

(1) CO_{2} = 200 mL: CO = 500mL

(2) CO_{2} = 350 mL: CO = 350mL

(3) CO_{2} = 0.0 mL: CO = 700mL

(4) CO_{2} = 300 mL: CO = 400mL

**Solution:**

CO_{2 }(g) + C (s) → 2CO (g)

Total volume = 700 ml = 0.7 L

0.5+x = 0.7

x = .2L = 200 mL

CO_{2} (g) = 0.5-0.2 = 300ml

CO (g) = 2x = 400 mL

Hence option (4) is the answer.

**9. An open vessel at 300 K is heated till ⅖ th of the air in it is expelled. Assuming that the volume of the vessel remains constant, the temperature to which the vessel is heated is :**

(1) 750 K

(2) 400 K

(3) 500 K

(4) 1500K

**Solution:**

At constant V and P, n_{1}T_{1} = n_{2}T_{2}

n_{1} = n

n_{2} = n-2n/5 = 3n/5

T_{1} = 300 K

300 n = (3n/5) T_{2}

T_{2} = 300×5/3 = 500 K

Hence option (3) is the answer.

**10. The density of 3M solution of sodium chloride is 1.252 g mL ^{-1}. The molality of the solution will be (molar mass, NaCl = 58.5 g mol^{-1})**

(1) 2.18 m

(2) 3.00 m

(3) 2.60 m

(4) 2.79 m

**Solution:**

Given Molar mass of NaCl = 58.5 g

M = 3 mol L^{-1}

Mass of weight W_{2} of NaCl in 1L solution W_{2} = 3×58.5 = 175.5g

Mass of L solution = V × d

= 1000 ×1.25 = 1250g

Mass of H_{2}O in solution (W_{1}) = 1250-175.5 = 1074g

m = W_{2}×1000/M_{w2} ×W_{1} = (175.5×1000)/58.5×1074.5 = 2.79m

Hence option (4) is the answer.

**11. 0.6 g of urea on strong heating with NaOH evolves NH _{3}. Liberated NH_{3} will combine with which of the following HCl solution?**

(1) 100 mL of 0.2 N HCl

(2) 400 mL of 0.2 N HCl

(3) 100 mL of 0.1 N HCl

(4) 200 mL of 0.2 N HCl

**Solution:**

NH_{2}CONH_{2} + 2NaOH → Na_{2}CO_{3} + 2NH_{3}

2 mole of urea ≡ one mole of NH_{3}

one mole of NH_{3} = one mole of HCl

So one mole of HCl = 2 mole of urea = 2×0.6/60 = 0.02 mol.

Hence option (1) is the answer.

**12. Calculate the mass of FeSO _{4}.7H_{2}O which must be added in 100 kg of wheat to get 10 PPM of Fe.**

**Solution:**

Ppm = (Mass of Fe/total mass)×10^{6}

Total mass = 100 kg = 100 × 1000 g

Mass of Fe = (ppm × total mass )/10^{6}

= 10× 100 × 1000/10^{6} = 1 g

Molecular mass of FeSO_{4}.7H_{2}O = 278

Mass of one Fe = 56 g

56 g of Fe → 278 g of FeSO_{4}.7H_{2}O

So 1 g of Fe → 278/56 = 4.96 g

Hence 4.96 g is the answer.

**13. Given a solution of HNO _{3} of density 1.4 g/mL and 63% w/w. Determine the molarity of HNO_{3} solution.**

**Solution:**

Density = mass/volume of solution

Volume = mass / density = 100g/1.4 g/ml = (100/1.4)ml

Molarity = no. of moles of solute /Volume of solution( l) = 1.4×1000/100 = 14 M

Hence 14 M is the answer.

**14. A transition metal M forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be**

(1) MCl_{2}

(2) MCl_{4}

(3) MCl_{5}

(4) MCl_{3}

**Solution:**

Given vapour density = 94.8

Vapour density = molecular mass/2

Molecular mass = 94.8×2 = 189.6

Given 74.75% chlorine.

So 74.75/100 * 189.6 = 141.72 g of chloride is there.

Then the number of atoms of chloride will be 141.72/35.5 =3.97 which is approximately 4.

So the formula of metal chloride will be MCl_{4}.

Hence option (2) is the answer.

**15. 10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration?**

(1) 0.57 M

(2) 5.7 M

(3) 11.4 M

(4) 1.14 M

**Solution:**

No. of moles of NaOH in 10 mL of 2 M solution = (10/1000)×2 = 0.02 mol

Number of moles of NaOH in 200 mL of 0.5M solution = (200/1000)×0.5 = 0.1 mol

Total number of moles of NaOH = 0.02+0.1 = 0.12 mol

Total volume = 10+200 = 210 mL = 0.210 L

Final concentration = 0.12/0.210 = 0.57 M

Hence option (1) is the answer.

**16. A 5.2 molal aqueous solution of methyl alcohol, CH _{3}OH, is supplied. What is the mole fraction of methyl alcohol in the solution?**

(1) 0.086

(2) 0.050

(3) 0.100

(4) 0.190

**Solution:**

We know mole fraction = moles of solute/(moles of solute + moles of solvent)

Let mass of water is 1 kg . Moles of CH_{3}OH is 5.2

X_{solute }= 5.2/(5.2+1000/18) = 5.2/(5.2+55.556) = 5.2/60.756 = 0.086

Hence option (1) is the answer.

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