Question

When $\text{20} \, \text{g}$ of naphthoic acid $\left(\text{(C}\right)_{\text{11}} \left(\text{H}\right)_{\text{8}} \left(\text{O}\right)_{\text{2}} \text{)}$ is dissolved in $\text{50} \, \text{g}$ of benzene
$\text{k}_{\text{f}} \, \text{=} \, \text{1} \text{.72 Kkg mol}^{- \text{1}}$ , a freezing point depression of $\text{2} \, \text{K}$ is observed. The van't Hoff factor (i) is

• A. 0.5
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

Actual molecular weight of naphthoic acid $\left(\text{(C}\right)_{\text{11}} \left(\text{H}\right)_{\text{8}} \left(\text{O}\right)_{\text{2}} \text{)} \, \text{=} \, \text{172}$
Molecular mass (calculated) $\text{=} \, \frac{\text{1000} \, \text{\times } \, \text{k}_{\text{f}} \, \text{\times } \, \text{w}}{\text{W} \, \text{\times } \, \text{ΔT}_{\text{f}}}$
$= \frac{\text{1000 \times } \, \text{1} \text{.72} \, \text{\times } \, \text{20}}{\text{50} \, \text{\times } \, \text{2}} = 344$
van't Hoff factor (i) = $\frac{\text{actual mol} \text{.wt} \text{.}}{\text{calculated mol} \text{.wt} \text{.}} \, \text{=} \, \frac{\text{172}}{\text{344}}$
$ \, =0.5$