Question

At 100°C the vapour pressure of a solution containing 6.5 g a solute in 100 g water is 732 mm. If K_b = 0.52, the boiling point of this solution will be ______.

Options

• 102°C

• 100°C

• 101°C

• 100.52°C

Answer 1

At 100°C the vapour pressure of a solution containing 6.5 g a solute in 100 g water is 732 mm. If K_b = 0.52, the boiling point of this solution will be 101°C.

Explanation:

Given: Mass of solute = 6.5 g

Mass of solvent (water) = 100 g = 0.1 kg

Vapour pressure of solution = 732 mm

Vapour pressure of pure water at 100°C = 760 mm

K_b = 0.52 K kg mol⁻¹

We know that

`(P_A^circ - P_A)/(P_A^circ) = chi_2`

`(760 - 732)/760 = chi_2`

`chi_2 = 28/190`

χ₂ = 0.03684    ...(i)

But `chi_2 = n_2/(n_1 + n_2) = n_2/n_1`

So, `0.03684 = n_2/n_1`    ...[From equation (i)]

⇒ n₂ = 0.03684 × n₁    ...(ii)

But `n_1 = "mass"/"molar mass" = 100/18 = 5.56`

Equation (ii) becomes,

n₂ ​= 0.03684 × 5.56 ≈ 0.205 mol

Molecular mass of solute `M = w_2/n_2 = 6.5/0.205 = 31.7` g/mol

Molality of solution `m = n_2/solvent = 0.205/0.1 = 2.05` mol/kg

We know that

ΔT_b = K_b × m

= 0.52 × 2.05 

= 1.07 K

T_b​ = 100 + ΔT_b

Boiling point = 100 + 1.07

≈ 101°C