Question

when 1.5g of a non volatile solute was dissolved in 90 g of benzene the boiling point of benzene raised from 353.23K to 353.93K. Calculate the molar mass of the solute.(Kb for benzene= 2.52K kg mol-1)

Answer 1

ΔT_b = K_b · m
where,
ΔT_b is the boiling point elevation, T_b (solution) - T_b (pure solvent),
K_b is the ebullioscopic constant,
​m is the molality of the solution
ΔT_b = 0.70 K
weight of solute = 1.5 g
weight of solvent= 90 g
molality = ​
$$\begin{gathered} \frac{\mathrm{weight}\mathrm{of}\mathrm{solute}}{\mathrm{molar}\mathrm{weight}\mathrm{of}\mathrm{solute}}\times \frac{1000}{\mathrm{weight}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{g}} \\ =\frac{1.5}{\mathrm{M}}\times \frac{1000}{90\mathrm{g}} \end{gathered}$$
Now,

$0.70\mathrm{K}=2.52\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}\times \frac{1.5}{\mathrm{M}}\times \frac{1000}{90}$

$\mathrm{M}=\frac{2.52\mathrm{K}\mathrm{kg}{{\mathrm{mol}}^{-}}^1\times 1.5\times 1000}{0.70\mathrm{K}\times 90}=60gmo{l}^{-1}$

Molecular weight of solute is 60 g mol^-1.