Question

The vapour pressure of pure benzene at certain temperature is $1$ bar. $A$ non-volatile, non-electrolyte solid weighing $2g$ when added to $39g$ of benzene (molar mass $78gmo{l}^{-1}$ ) yields solution of vapour pressure of $0.8$ bar. The molar mass of solid substance is

• A. 32
• B. 16
• C. 64
• D. 48

Answer 1

Answer: (B)

Step I Calculation of number of moles
$p_{\text{solution }}=p_{\text{benzene }}\times {\chi }_{\text{benzene }}$
$P_{\text{solution }}=0.8\text{ bar }$
$p_{C_6{H}_6}=1\text{ bar }$
${\chi }_{C_6{H}_6}=\frac{p_{\text{solution }}}{p_{C_6{H}_6}}=\frac{0.8bar}{1bar}$
${\chi }_{C_6{H}_6}=0.8$
${\chi }_{C_6{H}_6}=\frac{\text{ Moles of }{C}_6{H}_6}{\text{ Moles of }{C}_6{H}_6+\text{ Moles of solute }}$
${\chi }_{C_6{H}_6}=\frac{(39g)/78gmo{l}^{-1}}{(39g)/\left(78mo{l}^{-1}\right)+n_{\text{solute }}}$
$0.8=\frac{0.5}{0.5+n_{\text{selute }}}$
$0.4+0.8n$ solute $=0.5$
$0.8n$ solute $=0.5-0.4$
$n$ solute $=0.1/0.8$
$n$ solute $=\frac{1}{8}mol$
Step II Calculation of molecular mass of solute
Molecular mass $=\frac{\text{ Mass of solute }}{\text{ Number of moles of solute }}$
Molecular mass $=\frac{2.g}{(1/8)mol}=16gmo{l}^{-1}$