# 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. 1174 g/molB. 0.85 g/molC. 0.13 g/molD. 131 g/mol E. 95 g/mol

###### FREE Expert Solution

We’re being asked to determine the molar mass of an unknown of non-electrolytic solute present in a solution with a freezing point of –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:

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

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{imK}}}_{{\mathbf{f}}}}$

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:

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 unknown solute present.
Step 4: Calculate the molar mass of the solute.

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###### Problem Details

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. 1174 g/mol

B. 0.85 g/mol

C. 0.13 g/mol

D. 131 g/mol

E. 95 g/mol