Question

$1.22\text{g}$ of a monobasic acid is dissolved in $100\text{g}$ of benzene. Boiling point of solution increases by ${0.13}^{\circ }\text{C}$ with respect to pure benzene. Find the molecular mass of acid in benzene solvent (in $\text{u}$). Report your answer after dividing it by $100$ and round it off to the nearest integer.

$(K_b\text{of benzene}=2.6{\text{K kg mol}}^{-1}$)

• A. 2.0
• B. 2.00
• C. 2

Answer 1

Answer: (A), (B), (C)

Given,
B.P elevation $={0.13}^{\circ }$
$\text{Here, m = Mass of benzene}=100\text{g}=0.1\text{kg}$
$K_b\text{of benzene}=2.6{\text{K kg mol}}^{-1}={2.6}^{\circ }{\text{C kg mol}}^{-1}$

$\Delta T=i{K}_{b}m.....\left(i\right)$

Where $\Delta T$ is elevatioin in boiling point

$K_b$ is molal elevation constant

$i$ is van't hoff factor ($i\text{for benzene}=1$)

m is molality

$\text{Molality}=\frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$

Let the number of moles of acid in benzene = $x$
Equation $\left(i\right)$ becomes :

$\Rightarrow {0.13}^{\circ }\text{C}=\left(1\right)\left({2.6}^{\circ }{\text{C kg mol}}^{-1}\right)\times \frac{x}{0.1\text{kg}}$

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

$\Rightarrow x=1/200\text{moles}$
$\Rightarrow x=0.005\text{moles}$

$\text{moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\Rightarrow 0.005=\frac{1.22\text{g}}{\text{molecular weight}}$

$\Rightarrow \text{molecular weight}=\frac{1.22\text{g}}{0.005\text{moles}}$

$\Rightarrow \text{Molecular weight of acid}=\frac{244}{100}=2.44\text{u}\approx 2\text{u}$