Mole Concept MCQs-JEE Previous Year Questions With Solutions

The chapter contains important topics such as atomic and molecular mass, molar mass, gram atomic mass and gram molecular mass. ANAND CLASSES (A School Of Competitions) provides accurate solutions of Mole Concept MCQs.

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 6.023×10²³ particles.

1 mole of iodine atom= 6.023×10²³

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×10²³no.of Mg

Given 48g of Mg = 2×6.023×10²³

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 10²³

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 10²³

no. of chlorine atom = 6.023×10²³

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 10²³

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_xH_yO_z) is 6 : 1. If one molecule of the above compound ( C_xH_yO_Z ) contains half as much oxygen as required to burn one molecule of compound C_XH_Y completely to CO₂ and H₂O. The empirical formula of compound C_xH_yO_z is :

(1) C₂H₄O

(2) C₃H₄O₂

(3) C₂H₄O₃

(4) C₃H₆O₃

Solution:

Given the ratio of mass percent of C and H of an organic compound (C_xH_yO_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_xH_yO_Z ) contains half as much oxygen as required to burn one molecule of compound C_XH_Y completely to CO₂ and H₂O

C_XH_y + O₂ → xCO₂+ (y/2)H₂O

Put y = 2x in above equation

C_XH_2x + O₂ → xCO₂+ xH₂O

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 C₂H₄O₃.

Hence option (3) is the answer.

3. The concentrated sulphuric acid that is peddled commercially is 95% H₂SO₄ 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% H₂SO₄ means 100 gm contain 95 gm H₂SO₄

Mass of solute = (95/100)×1834

Molecular weight of H₂SO₄ = 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 O₂ = w

Mass of N₂ = 4w

Molecules of O₂ = w/(32×N_A)

Molecules of N₂ = 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 CH₃COOH 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 (O₂) is :

(Atomic mass :C =12, O =16 and Avogadro’s constant N_A = 6.0 * 10²³ mol^-1)

(1) 3 : 1 : 1

(2) 1 : 1 : 2

(3) 3 : 1 : 2

(4) 1 : 1 : 1

Solution:

Molar mass of O₃ = 48

Given 16 g O₃ . So no. of moles of O₃ = 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 O₂ = 32

Given 32g O₂

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 CO₂ (g) is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5 litre of CO₂ (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) CO₂ = 200 mL: CO = 500mL

(2) CO₂ = 350 mL: CO = 350mL

(3) CO₂ = 0.0 mL: CO = 700mL

(4) CO₂ = 300 mL: CO = 400mL

Solution:

CO₂ (g) + C (s) → 2CO (g)

Total volume = 700 ml = 0.7 L

0.5+x = 0.7

x = .2L = 200 mL

CO₂ (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, n₁T₁ = n₂T₂

n₁ = n

n₂ = n-2n/5 = 3n/5

T₁ = 300 K

300 n = (3n/5) T₂

T₂ = 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 W₂ of NaCl in 1L solution W₂ = 3×58.5 = 175.5g

Mass of L solution = V × d

= 1000 ×1.25 = 1250g

Mass of H₂O in solution (W₁) = 1250-175.5 = 1074g

m = W₂×1000/M_w2 ×W₁ = (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 NH₃. Liberated NH₃ 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:

NH₂CONH₂ + 2NaOH → Na₂CO₃ + 2NH₃

2 mole of urea ≡ one mole of NH₃

one mole of NH₃ = 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 FeSO₄.7H₂O which must be added in 100 kg of wheat to get 10 PPM of Fe.

Solution:

Ppm = (Mass of Fe/total mass)×10⁶

Total mass = 100 kg = 100 × 1000 g

Mass of Fe = (ppm × total mass )/10⁶

= 10× 100 × 1000/10⁶ = 1 g

Molecular mass of FeSO₄.7H₂O = 278

Mass of one Fe = 56 g

56 g of Fe → 278 g of FeSO₄.7H₂O

So 1 g of Fe → 278/56 = 4.96 g

Hence 4.96 g is the answer.

13. Given a solution of HNO₃ of density 1.4 g/mL and 63% w/w. Determine the molarity of HNO₃ 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) MCl₂

(2) MCl₄

(3) MCl₅

(4) MCl₃

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 MCl₄.

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, CH₃OH, 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 CH₃OH 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.