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 (ΔT_{f})** is given by:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{pure}\mathbf{}\mathbf{solvent}}{\mathbf{-}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{solution}}}$

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)

**K _{f}** = freezing point depression constant (in ˚C/m)

Recall that the ** molality of a solution** is given by:

$\overline{){\mathbf{molality}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}}{\mathbf{kg}\mathbf{}\mathbf{solvent}}}$

**For this problem, we need to do the following:**

**Step 1:**** **Calculate for ΔT_{f}.

* Step 2:* Determine the molality of the solution.

* Step 3:* Calculate the moles of the unknown solute present.

0.528 grams of an unknown non-electrolytic solute were dissolved in 50.0 g of water (K_{f}_{p} = 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

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