Question

Calculate the boiling point of a solution containing 0.61 g of benzoic acid in 50 g of carbon-disulphide assuming 84% dimerisation of the acid. The boiling point and K_b of CS₂ are 46.2°C and 2.3 K kg mol⁻¹ respectively.

Answer 1

Given: Mass of benzoic acid = 0.61 g

Molar mass of benzoic acid = 122 g/mol

Mass of carbon disulphide = 50 g = 0.050 kg

Boiling point of pure CS₂ = 46.2°C

K_b​ for CS₂ = 2.3 K kg mol⁻¹

Degree of dimerisation = 84% = 0.84

Molality (m) = `"moles of solute"/"mass of solvent in kg"`

= `(0.61//122)/0.050`

= 0.1 mol/kg

Benzoic acid dimerises in a non-polar solvent. The effective number of particles changes as follows:

`i = 1 − alpha/2` (for dimerisation)

`i = 1 − 0.84/2`

= 1 − 0.42

= 0.58

ΔT_b = i . k_b . m

= 0.58 × 2.3 × 0.1

= 0.1334°C

`T_"boiling (solution)" = T_"boiling(solvent)" + Delta T_b`

= 46.2 + 0.1334

= 46.33°C

∴ The boiling point of the solution is 46.33°C.

Answer 2

The degree of association of benzoic acid

`alpha = 84/100 = 0.84`

In solution,

	\[\ce{2C6H5COOH <=> (C6H5COOH)2}\]
Initially	1 mol                                         -
At equilibrium	(1 − 0.84) mol            `0.84/2 = 0.42` mol

 ∴ Number of moles in solution = 1 − 0.84 + 0.42

= 0.58

∴ `i = "No. of moles in solution"/"No. of moles added"`

= `0.58/1`

= 0.58

Number of moles of solute (benzoic acid) present in solution

= `0.61/122`    ...(∵ Normal molecular mass of C₆H₅COOH = 122)

= 5 × 10⁻³

Mass of solvent (CS₂) = 50g = 0.05 kg

∴ Molality (m) of the solution = `(5 xx 10^-3)/0.05`

= 0.1

According to the modified equation for elevation of boiling point,

ΔT_b = i . k_b . m

= 0.58 × 2.3 × 0.1

= 0.1334°C

Hence, the boiling point of the solution = 46.2 + 0.1334

= 46.333°C