Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of NaNO₃. The molal freezing- point- depression constant of water is 1.86 °C/m. A) 0.1023 B) 0.0286 C) - 0.05627 D) - 0.1023 E) - 0.2046

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the solution:
molality (m) = moles of solute / mass of solvent (in kg)
Since we are given the concentration of the solution in molarity (M), we first need to convert it to moles of solute per kilogram of solvent:
moles of solute = (0.05500 mol/L) x (1 L / 1000 mL) x (22.99 g NaNO3 / 1 mol NaNO3) = 0.00126495 mol
mass of solvent = 1000 g - 55 g (mass of solute) = 945 g = 0.945 kg
m = 0.00126495 mol / 0.945 kg = 0.001338 mol/kg

Next, we can use the formula for freezing-point depression:
ΔTf = Kf x m
where ΔTf is the change in freezing point, Kf is the molal freezing-point depression constant, and m is the molality of the solution.

We are given Kf, so we can plug in the values to solve for ΔTf:
ΔTf = (1.86 °C/m) x (0.001338 mol/kg) = 0.00248668 °C

Finally, we can calculate the freezing point (Tf) of the solution:
Tf = 0°C - ΔTf = 0°C - 0.00248668°C = -0.0025°C

Since the freezing point is below 0°C, we round down to -0.005°C or -0.01°C. The only answer choice that is closest to either of these values is E (-0.2046).

Calculate the Freezing Point (0°C) of a 0