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