Question

The vapour pressure of pure benzene at a certain temperature is $640mmHg$. Anon-volatile, non-electrolyte solid weighing $2.175g$ is added to $39.0g$ of benzene. The vapour pressure of the solution is $600mmHg$. Molecular weight of the non-electrolyte is

• A. $92.05gmo{l}^{-1}$
• B. $60.08gmo{l}^{-1}$
• C. $72.08gmo{l}^{-1}$
• D. $65.25gmo{l}^{-1}$

Answer 1

Answer: (D)

By Raoult's law, $\frac{p^{\circ }-p_s}{p^{\circ }}={\chi }_{\text{solute }}$

$=\frac{n_{\text{solute }}}{n_{\text{solute }}+n_{\text{solvent }}}$

Given, $p^{\circ }=640mmHg$ ,

$p_s=600mmHg$

$n_{\text{solute }}=\frac{2.175}{m_1}$

$(m_1=$ molecular weight of solute $)$

$n_{\text{solvent }}=\frac{39}{78}=0.5$

$\therefore \frac{40}{640}=\left(\frac{2.175/m_1}{\frac{2.175}{m_1}+0.5}\right)$

$\Rightarrow \frac{\left(2.175/m_1\right)+0.5}{\left(2.175/m_1\right)}=16$

$\frac{0.5}{\left(\frac{2.175}{m_1}\right)}=15$

$\Rightarrow \frac{2.175}{m_1}=\frac{0.5}{15}$

$\therefore m_1=65.25gmo{l}^{-1}$