Question

Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol −1 .

Answer 1

Answer :

Molecular weight of $C^{6}H^{8}O^{6}$ = 72 + 8 + 96 = 176 g $mol^{-1}$

Δ$T_{f} = K_{f}$ × m = $K_{f} × \frac{W_{B}}{M_{B}}$ × $\frac{1000}{W_{A}}$

⇒ 1.5 = 3.90 × $\frac{W_{B}}{176} × \frac{1000}{75}$

⇒ $W_{B} = \frac{1.5 × 176 × 75}{3.90 × 1000} = \frac{19800}{3900}$

⇒ 5.076 g

Answer 2

Given,
Mass of acetic acid (solvent), $w_1$= 75 g
Lowering of melting point, ΔTf = 1.5 K
Kf = 3.9 K kg/mol
Molar mass of ascorbic acid $C_6H_8O_6$ ,M= 12 × 6 + 8 × 1 + 16 × 6 = 72 + 8 + 96 =176g/mol
let the weight of ascorbic acid (solute),$w_2$

we know, $\Delta{T}=\frac{K\times1000\times w_2}{M\times w_1}$
1.5 = (3.9 × 1000 × w₁)/(176 × 75)
1.5 = (3900w₁)/(176 × 75)
1.5 × 176 × 75 = 3900w₁
w₁ = 1.5 × 176 × 75/3900 = 5.08g

hence, weight of ascorbic acid = 5.08g