A pure solvent freezes at 43.4°C.The freezing temperature is 40.7°C when 17.5 g of CO(NH₂)₂ is dissolved in 1.00 × 10² g of the solvent.Calculate the molal freezing point constant of the solvent. A)0.027 °C/m B)0.343 °C/m C)0.93 °C/m D)13.9 °C/m E)14.9 °C/m

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Quizplus · Accepted Answer

The freezing point depression is ΔTf = 43.4°C - 40.7°C = 2.7°C. The molality (m) of the solution is calculated using the formula: m = moles of solute / kg of solvent. Moles of CO(NH2)2 = 17.5 g / 60.06 g/mol = 0.291 moles. Molality = 0.291 moles / 0.1 kg = 2.91 m. The molal freezing point constant (Kf) is calculated using ΔTf = Kf * m, so Kf = ΔTf / m = 2.7°C / 2.91 m = 0.93 °C/m.

A Pure Solvent Freezes at 43