Question

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

• A. $92.05 \,g \,mol^{-1}$
• B. $60.08\,g \,mol^{-1}$
• C. $72.08\,g \,mol^{-1}$
• D. $65.25\,g \,mol^{-1}$

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, $p^{\circ}=640\, mm \,Hg$ ,

$ p_{ s }=600 \,mm \,Hg$

$n_{\text {solute }}=\frac{2.175}{m_{1}}$

$\left(m_{1}=\right.$ 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 $

$\Rightarrow \frac{2.175}{m_{1}}=\frac{0.5}{15}$

$\therefore m_{1}=65.25\, g\, mol ^{-1}$