Question

1.22 g of a monobasic acid is dissolved in 100 g of benzene. The boiling point of the solution increases by 0.13${}^{\circ }C$ with respect to pure benzene. Find the molecular mass of aid in benzene solvent (in u). Report your answer after dividing it by 100 and Round it off to the nearest integer. ($K_b$ of benzene = 2.6 K kg mol${}^{-1}$).

Answer 1

B.P. elevation= ${0.13}^{o}C$ Here, $m=$ Mass of benzene=$100g=0.1kg$

$K_b$ of benzene= $2.6kKgmo{l}^{-1}=2.6C^{o}Kgmo{l}^{-1}$

$\Delta T=i{K}_{b}m$

$\Rightarrow {0.13}^{o}C=\left(1\right)\left(2.6C^{o}Kgmo{l}^{-1}\right)\times \frac{x}{\left(0.1\right)}kg$

$\Rightarrow 0.13=2.6\times \frac{x}{0.1}$

$\Rightarrow x=\frac{0.13\times 0.1}{2.6\times 1000}\times 10$ moles

$\Rightarrow x=\frac{1}{200}$ moles

$\Rightarrow x=0.005$ moles

Molecular weight= $\frac{Weighting}{Moles}$

$\Rightarrow \frac{1.22}{0.005}=\frac{122}{5}\times \frac{1000}{100}$

$\Rightarrow 24.4\times 10=244g/mol$

Molecular weight= $\frac{244}{100}=2.44u=2u$