When 1.5 g of a non-volatile solute is dissolved in 30 g of a solvent, the boiling point of the solvent is raised by 2K. Calculate the molar mass of the solute, given that \(K_b\) for the solvent is 1.85 \(K kg mol^{-1}\).

46.25 \(g mol^{-1}\)

Given,

\(W_A = 30 g\)

\(W_B = 1.5 g\)

\(\Delta T_b = 2K\)

\(K_b = 1.85 \text{K Kg mol}^{-1}\)

We know,

\(M_B = \frac{K_b × W_B × 1000}{\Delta T_b × W_A}\)

\(⇒ M_B = \frac{1.85 × 1.5 × 1000}{2 × 30}\)

\(⇒ M_B = \frac{2775}{60}\)

\(⇒ M_B = 46.25 \text{ g mol}^{-1}\)