Question

$0.6\, mL$ of acetic acid $(CH_{3}COOH)$, having density $1.06\, g\, mL^{-1}$, is dissolved in $1$ litre of water. The depression in freezing point observed for this strength of acid was $0.0205^{\circ}C$. The van’t Hoff factor is

• A. $1$
• B. $2$
• C. $5$
• D. $7$

Answer 1

Answer: (A)

Mass of acetic acid
$w_{2} =$ Volume $\times$ Density $ = 0.6 \times 1.06 = 0.636\, g$ We know,
$\Delta T = i \times K_{f} \times \frac{w_{2} \times 1000}{M_{2} \times w_{1}}$ ...(i)
Given, $\Delta T = 0.0205,\, K_{f} = 1.86\, K\, kg\,$ mol $^{-1}, w_{2} = 0.636\, g$
$M_{2} = 60 \,g$ mol $^{-1}, w_{1} = 1000\, g$ (Mass of 1 litre water) Putting these values in eqn. (i), we get
$0 .0205 = i \times 1.86 \times \frac{0.636 \times 1000}{60 \times 1000}$
$i = 1.03397$