Question

The boiling point of benzene is $353.23K$. When $1.80g$ of a non-volatile solute was dissolved in $90g$ of benzene, the boiling point is raised to $354.11K$. Calculate the molar mass of the solute. $K_b$ for benzene is $2.53Kkgmo{l}^{-1}$

Answer 1

Elevation in boiling point

We know that, the elevation in the boiling point $\Delta T_b=T_b-T_b^0$
By putting the value, we get;
$\Delta T_b=354.11K-353.23K=0.88K$

Molar mass of solute

We know that;
$\Delta T_b=K_b\times \text{molality(m)}$

We also know that molality$=\frac{\text{mass of solute}}{\text{Molar mass of solute}\times \text{mass of solvent(kg)}}$
$\Delta T_b=K_b\times \frac{\text{mass of solute}\left(m_2\right)}{\text{molar mass of solute}\left(M_2\right)\times \text{mass of solvent}\left(m_1\right)}$

So,
$M_2=K_b\times \frac{\text{mass of solute}\left(m_2\right)}{\Delta T_b\times \text{mass of solvent}\left(m_1\right)}$

$M_2=\frac{2.53Kkgmo{l}^{-1}\times 1.8g}{0.88K\times 90\times {10}^{-3}Kg}=57.5gmo{l}^{-1}$

Hence, the molar mass of the solute $M_2=57.5gmo{l}^{-1}$.