Question

Boiling point of benzene is $353.23\, K$. When $1.8\, g$ of non-volatile solute is dissolved in $90\, g$ of benzene. Then boiling point is raised to $354.11\, K$. Given $K_b$ (benzene) = $2.53 \, kg\, mol^{-1}$. The molecular mass of non-volatile substance is

• A. $58 \,g\, mol^{-1}$
• B. $120 \,g\, mol^{-1}$
• C. $116 \,g\, mol^{-1}$
• D. $60 \,g\, mol^{-1}$

Answer 1

Answer: (A)

$T_{b}^{\circ}=353.23 K,\, W_{B}=1.8\, g,$

$W_{A}=90\, g,\, T_{b}=354.11\, K,$

$K_{b}=2.53\, kg\, mol^{ -1}$

$\Delta T_{b}=T_{b}-T_{b}^{\circ}=354.11-353.23=0.88\, K$

$M_{B}=\frac{W_{B}\times K_{b}\times1000}{\Delta T_{b}\times W_{A}}=\frac{1.8\times2.53\times1000}{0.88\times90}$

$=57.5\approx 58\, g\, mol^{ -1}$