Question

A solution is prepared by mixing 0.3355 moles of NaNO3 with 235 g of water. Its density is 1.0733 g mL^-1. If Kf= 1.86 C m^-1 and the solute is completely ionized what is the expected freezing point.

Answer 1

To find the expected freezing point of the solution, we need to calculate the molality and use the equation ΔTf = Kf * m. The molality (m) is the number of moles of solute per kilogram of solvent. We can then calculate the freezing point depression, ΔTf, and subtract it from the freezing point of pure water to get the expected freezing point of the solution.

Step-by-step explanation:

To find the expected freezing point, we first need to calculate the molality (m) of the solution. The molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is NaNO3 and the solvent is water.

To calculate m, we divide the moles of NaNO3 by the mass of water in kilograms:

m = 0.3355 moles NaNO3 / (235 g water / 1000 g/kg) = 1.429 mol/kg

Next, we can use the equation ΔTf = Kf * m to calculate the freezing point depression. ΔTf represents the change in freezing point of the solvent, Kf is the molal freezing point depression constant for the solvent (1.86 °C/m for water), and m is the molality of the solution. Plugging in the values, we get:

ΔTf = 1.86 °C/m * 1.429 mol/kg = 2.66094 °C

The expected freezing point is the freezing point of pure water minus ΔTf:

Freezing point = 0 °C - 2.66094 °C = -2.66094 °C