Question

The vapour pressure of pure benzene at certain temperature is $1$ bar. $A$ non-volatile, non-electrolyte solid weighing $2\, g$ when added to $39\, g$ of benzene (molar mass $78\, g\,mol ^{-1}$ ) 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
$p_{\text {solution }}=p_{\text {benzene }} \times \chi_{\text {benzene }}$
$P_{\text {solution }}=0.8 \text { bar }$
$p_{C_{6} H_{6}}=1 \text { bar }$
$\chi_{C_{6} H_{6}}=\frac{p_{\text {solution }}}{p_{C_{6} H_{6}}}=\frac{0.8 bar }{1 bar }$
$\chi_{C_{6} H_{6}}=0.8$
$\chi_{C_{6} H_{6}}=\frac{\text { Moles of } C _{6} H _{6}}{\text { Moles of } C _{6} H _{6}+\text { Moles of solute }}$
$\chi_{C_{6} H_{6}}=\frac{(39\, g ) / 78\, gmol ^{-1}}{(39\, g ) /\left(78\, mol ^{-1}\right)+n_{\text {solute }}}$
$0.8=\frac{0.5}{0.5+n_{\text {selute }}}$
$0.4+0.8 n$ solute $=0.5$
$0.8 n$ 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 $=\frac{2. g}{(1 / 8) mol }=16\, g\,mol ^{-1}$