# Solutions Chemistry Class 12 Chapter Solution Test Paper – 2

Solutions Class 12 Chapter Solution Test Paper

**Ans 1.**

**Ans 2.** The solute is 50% dissociated

** **

**Ans 3.**

**Ans 4.**

**Ans 5. **According to Raoult’s law,** **

(p^{0} – p) / p^{0} = (w × M) / (m × W)

p^{0} → is Vapour pressure of pure water

p → is vapour pressure of the solvent

w → is weight of solute

m → is the molecular weight of solute

W → is the molecular weight of solvent

(32 – 31.84) / 32 = (10 × 18) / (m × 200)

m = 180 g/mol.

Solutions Class 12 Chapter Solution Test Paper

**Ans 6. **p^{0} = 0.850 bar = Vapour pressure of pure benzene.

p = 0.845 bar = Vapour pressure of solution.

w_{1} = 39 g = Mass of benzene

w_{2} = 78 g = Mass of solute

M_{1} = 78 g/mol = Molar mass of benzene.

M_{2} = Molar mass of solute.

(p^{0} – p)/p = w_{2 }× M_{1 }**/ **w_{1} × M_{2}

(p^{0} – p) = 0.850 – 0.845 = 0.005 bar

w_{2} × M_{1} = 0.5 × 78 = 39

w_{1} × M_{2} = 39 × M_{2}

0.005 = 39/39 × M_{2}

0.005/0.850 = 1/M_{2}

M_{2} = 1 × 0.850 / 0.005 = 170 g/mol

**Ans 7.**

**Ans 8.** Moles of glucose = 18 g/ 180 g mol^{–1} = 0.1 mol

No of kg of solvent = 1 kg

therefore, molality of glucose solution = 0.1 mol kg^{-1}

For water, change in boiling point

ΔT_{b} = K_{b} × m = 0.52 K kg mol^{–1} × 0.1 mol kg^{–1} = 0.052 K

Since water boils at 373.15 K at 1.013 bar pressure.

The boiling point of solution will be 373.15 + 0.052 = 373.202 K.

**Ans 9. **80.2 g of water contains = 0.520 g

1000 g of water contains = 0.52 × 1000/80.2 = 6.484 g of Glucose.

No. of moles of Glucose = 6.484/180 = 0.036

increase in boiling point of water = K_{b} × m,

where’s K_{b} = 0.52, m = 0.036.

So ∆T_{b} = 0.52 × 0.036 = 0.0187

So boiling point of Solution

= 100° + 0.0187° = 100.0187°C

= 273 + 100.0187 = 300.0187 K

**Ans 10. ** K_{b} for benzene = 2.53 K KJ/mole

Mass of solute = W_{2} = 1.8 g = 1.8 × 10^{-3} kg

Mass of Solvent = W1 = 90g = 90 × 10^{-3}

Boiling point of solution = T_{b} = 354.11 K

Boiling point of pure solvent = T_{b}’ = 353.23 K

**Ans 11. **Given

**Ans 12. **We know

**Ans 13.** CCl_{3}COOH → CCl_{3}COO^{− }+ H^{+}

ΔT_{b }= (100.18 + 273) − (100 − 273) = 0.18K

Molality is 1.

Hence, ΔT_{b} = iK_{b}m = iK_{b }× 1

i = ΔT_{b }/ K_{b}

i = 0.18 / 0.512

i = 0.351

**Ans 14. **We know

ΔT_{f} = i K_{f} m

2.93 = 1.86 × 5.61 × i

i = 1.86 × 5.61 / 2.93

i = 0.28

**Ans 15. **We know

ΔT_{f} = i K_{f} m

Freezing point of solution = 273 – 2.79 = 270.21 K

**Ans 16. **1 molar solution means that 1 gram mole of KCI is dissolved in one litre of the solution.

Mass of 1L of solution = V × d = 1000 × 1.04 = 1040 g

Mass of one mole KCI = 39 + 35.5 = 74.5 g

Mass of solvent =1040 – 74.5 = 965.5 g = 0.965 kg

Molality of solution (m) = No. of moles of solute / Mass of solvent in kg

= 1/ 0.965 mol/kg = 1.0357 m

KCl dissociates as: KCl → K^{+} + Cl^{−}

∴ Number of particles after dissociation = 2

∴ van’t Hoff factor, i = 2

Now ΔT_{b} = i × K_{b} × m = 2 × 0.52 × 1.0357 = 1.078 ^{∘}C

∴ Boiling point of the solution = 100 + 1.078 = 101.078 ^{∘}C

**Ans 17. **As NaCl undergoes complete dissociation

NaCl → Na^{+} + Cl^{−}

One mole of NaCl will give 2 particle and the value of i will be equal to 2.

ΔT_{f} = i k_{f} m

K_{f} = 1.86 K m^{−1},

ΔT_{f} = 3K, i=2

∴ 3 = 2 × 1.86 × m

∴ m = 0.806 mol/kg

Amount of NaCl = 0.806 × 58.5 = 47.151 = 0.806 × 58.5 = 47.151 g/kg

**Ans 18. **

**Ans 19.**

**Ans 20.**