Question

The addition of 0.643 g of a compound to 50 mL of benzene (density $0.879g{mL}^{-1}$) lowers the freezing point from $5.51$ to ${5.03}^{o}C$. If $K_f$ for benzene is $5.12K/m$, calculate the molecular weight of the compound.

Answer 1

According to depression in freezing point method

$M_2=\frac{1000\times K_f}{\Delta T_f}\times \frac{W_2}{W_1}$

Given that

$K_f\to$ molar depression constant of benzene
$=5.12Kkg{mol}^{-1}$

$W_2=$ weight of compound $=0.643g$

$W_1=$ weight of benzene $=$ Volume $\times$ Density
$=50\times 0.879=43.95g$

$\Delta T_f\to$ depression in freezing point $=5.51-5.03={0.48}^{o}C$

Therefore, on substitution

$M_2=\frac{1000\times 5.12\times 0.643}{0.48\times 43.95}=156.06g/mol$