Boiling point of benzene is $353.23 \mathrm{~K}$. When $1.8 \mathrm{~g}$ of non-volatile solute is dissolved…

Boiling point of benzene is $353.23 \mathrm{~K}$. When $1.8 \mathrm{~g}$ of non-volatile solute is dissolved in $90 \mathrm{~g}$ of benzene. Then boiling point is raised to $354.11 \mathrm{~K}$. Given $K_b$ (benzene) $=2.53 \mathrm{~kg} \mathrm{~mol}^{-1}$. The molecular mass of non-volatile substance is
  1. $58 \mathrm{~g} \mathrm{~mol}^{-1}$
  2. $120 \mathrm{~g} \mathrm{~mol}^{-1}$
  3. $116 \mathrm{~g} \mathrm{~mol}^{-1}$
  4. $60 \mathrm{~g} \mathrm{~mol}^{-1}$

Solution

$T_b^{\circ}=353.23 \mathrm{~K}, W_B=1.8 \mathrm{~g}$, $W_A=90 \mathrm{~g}_{\prime} T_b=354.11 \mathrm{~K}$, $K_b=2.53 \mathrm{~kg} \mathrm{\textrm {mol } ^ { - 1 }}$ $\Delta T_b=T_b-T_b{ }^{\circ}=354.11-353.23=0.88 \mathrm{~K}$ $M_B=\frac{W_B \times K_b \times 1000}{\Delta T_b \times W_A}=\frac{1.8 \times 2.53 \times 1000}{0.88 \times 90}$ $=57.5 \approx 58 \mathrm{~g} \mathrm{~mol}^{-1}$

Asked in: NEET 2012 (Mains)

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