Molar Mass of Solute from Boiling Point Elevation

The Eq. (2.13) is $\Delta T_b = K_b m$.

Suppose we prepare a solution by dissolving $\mathrm{W_2}$ g of solute in $\mathrm{W_1}$ g of solvent. The moles of solute in $\mathrm{W_1}$ g of solvent are $\dfrac{W_2}{M_2}$, where $M_2$ is the molar mass of the solute. The mass of the solvent, converted from grams to kilograms, is $\dfrac{W_1}{1000}$ kg.

Recall the expression for molality, $m$:

Substitution of this value of $m$ in Eq. (2.13) gives

Hence,

$M_2 = \frac{1000\,K_b W_2}{\Delta T_b W_1} \qquad \text{...(2.15)}$ …