Question

The Kf for acetic acid is 3.90 oC/m and its freezing point is 16.6 oC. Calculate the freezing point of a 1.25 M solution of benzoic acid, C7H6O2 in acetic acid. The density of this solution is 1.067 g/mL. The density of pure acetic acid is 1.049 g/mL.

Answer 1

Sol:-

Given, Moarity of the solution = 1.25 M means

1.25 moles of C7H6O2 i.e benzoic acid present in 1000 mL of solution .

also density = mass/ volume

therefore

mass of solution = density of solution x volume of solution

mass of solution = 1.067 g/mL x 1000 mL

mass of solution = 1067 g

also number of moles = given mass / molar mass

so

mass of benzoic acid = moles of benzoic acid x molar mass of benzoic acid

mass of benzoic acid = 1.25 mol x 122 g/mol = 152.5 g

Now

mass of solution = mass of benzoic acid + mass of acetic acid

mass of acetic acid = mass of solution -  mass of benzoic acid

mass of acetic acid = 1067 g - 152.5 g

mass of acetic acid = 914.5 g = 0.9145 kg

Now

Molality = number of moles of benzoic acid / mass of solvent i.e acetic acid in kg

Molality = 1.25 mol / 0.9145 kg

Molality = 1.37 mol/kg = 1.37 m

also

delta T_f = k_f x m

here k_f = molal depression constant = 3.90 ⁰C/m (given)

m= molality of the solution = 1.37 m and

delta T_f = change in freezing point = ?

delta T_f = 3.90 ⁰C/m x 1.37 m

delta T_f = 5.343 ⁰C

also delta T_f = T₀ - T_f

here , T₀ = freezing point of pure solvent i.e acetic acid and

T_f =  freezing point of Benzoic acid solution.

5.343 ⁰C = 16.6 ⁰C  - T_f

T_f =  16.6 ⁰C - 5.343 ⁰C

T_f = 11.26 ⁰C