2 g nonelectrolyte solute when dissolved in 50 g water increases its boiling point by 0.13 K.
Calculate the molar mass of solute if molal elevation constant of water is 0.52 K kg mol$^{-1}$

The elevation in boiling point is given by the formula:

$\Delta T_b = K_b \times m$

where $m$ is the molality of the solution, which can be expressed as:

$m = \dfrac{w_2 \times 1000}{M_2 \times w_1}$

Substituting the expression for molality into the boiling point elevation formula gives:

$\Delta T_b = \dfrac{K_b \times w_2 \times 1000}{M_2 \times w_1}$

Rearranging to solve for the molar mass of the solute, $M_2$:

$M_2 = \dfrac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1}$

Substituting the given values ($K_b = 0.52$ K kg mol$^{-1}$, $w_2 = 2$ g, $\Delta T_b = 0.13$ K, $w_1 = 50$ g):

$M_2 = \dfrac{0.52 \times 2 \times 1000}{0.13 \times 50}$

$M_2 = 4 \times 40 = 160$ g/mol