A solution containing 2.60 g of a non-volatile, non-electrolyte solute in 200 g of water boils at 100.130°C at 1 atm. What is the molar mass of the solute? [() = 0.52 K-kg/mol]

Question

A solution containing 2.60 g of a non-volatile, non-electrolyte solute in 200 g of water boils at 100.130°C at 1 atm. What is the molar mass of the solute? [$K_b$($H_2O$) = 0.52 K-kg/mol]

Prashant Kotian · Accepted Answer

Correct answer: A. Elevation of boiling point ΔT_b = T_boiling - T_pure = 100.130°C - 100.000°C = 0.130°C. Using the formula: ΔT_b = * m => 0.130 = 0.52 * (w_solute / M_solute) * (1000 / W_solvent) => 0.130 = 0.52 * (2.60 / M) * (1000 / 200) => 0.130 = 0.52 * (2.60 / M) * 5 = 6.76 / M => M = 6.76 / 0.130 = 52.0 g/mol.

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