A molecular solute with a molar mass of 50.0 g mol^-1 is dissolved in 500 g of water and the resulting solution has a boiling point of 101.53^°C. How many grams of solute were in the solution$\Delta$K_b = 0.512 °C m^-¹

Question

Quizplus · Accepted Answer

The boiling point elevation formula is ΔT_b = i * K_b * m. Here, ΔT_b = 101.53°C - 100°C = 1.53°C, K_b = 0.512°C m^-1, and i = 1 for a molecular solute. Solving for m (molality), m = ΔT_b / K_b = 1.53 / 0.512 = 2.988 mol/kg. Since the solvent is 0.5 kg of water, moles of solute = 2.988 * 0.5 = 1.494 mol. Mass of solute = moles * molar mass = 1.494 * 50 = 74.7 grams, approximately 75 grams.

A Molecular Solute with a Molar Mass of 50 Δ\DeltaΔ Kb = 0

A Molecular Solute with a Molar Mass of 50 $\Delta$ K_b = 0