Chemistry Practice Problems Freezing Point Depression Practice Problems Solution: 0.528 grams of an unknown non-electrolytic solute ...

Solution: 0.528 grams of an unknown non-electrolytic solute were dissolved in 50.0 g of water (Kfp = 1.86 K/m). If the freezing point of the solution was -0.150 °C, determine the molar mass of the unknown solute.A. 64 g/molB. 0.131 g/molC. 1310 g/molD. 13.1 g/molE. 131 g/mol

Problem

0.528 grams of an unknown non-electrolytic solute were dissolved in 50.0 g of water (Kfp = 1.86 K/m). If the freezing point of the solution was -0.150 °C, determine the molar mass of the unknown solute.

A. 64 g/mol
B. 0.131 g/mol
C. 1310 g/mol
D. 13.1 g/mol
E. 131 g/mol

Solution

We’re being asked to determine the molar mass of an unknown solute that was dissolved in 50.0 g of water. The freezing point of the solution is 0.150 °C.


Recall that the freezing point of a solution is lower than that of the pure solvent and the change in freezing point (ΔT­f) is given by:


ΔTf=Tf, pure solvent-Tf, solution


The change in freezing point is also related to the molality of the solution:


ΔTf=imKf


where: 

i = van’t Hoff factor

m = molality of the solution (in m or mol/kg)

Kf = freezing point depression constant (in ˚C/m)


Recall that the molality of a solution is given by:


molality=moles solutekg solvent


For this problem, we need to do the following:

Step 1: Calculate for ΔTf.

Step 2: Determine the molality of the solution.

Step 3: Calculate the moles of the solute present.
Step 4: Calculate the molar mass of the solute.


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