Can someone me ? 13. the boiling point of a solvent is elevated by 2.4 °c when the solute concentration is 3.1 m. what is kb? what is the freezing-point depression of a solution that contains 0.705 mol of a nonelectrolyte solute in 5.02 kg of water? (kf = 1.86 °c/m)

Answers:

0.77 °C·kg·mol⁻¹; 0.261 °C

Step-by-step explanation:

13. a. Boiling point elevation

The formula for boiling point elevation ΔTb is

ΔTb = Kb·b     Divide each side by b

Kb = ΔTb/b

Kb = 2.4/3.1

Kb = 0.77 °C·kg·mol⁻¹

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13. b. Freezing point depression

The formula for freezing point depression is

b = moles of solute/kilograms of solvent

b = 0.705/5.02

b = 0.1404 mol/kg