We’re being asked to **determine the molar mass of an unknown solute** present in **50.0 g benzene** to get a change in freezing point of** 0.41˚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:* Determine the molality of the solution.

* Step 2:* Calculate the moles of the unknown solute.

* Step 3:* Calculate the molar mass of the unknown solute.

When 0.81 g of an unknown non-electrolyte is dissolved in 50.0 g of benzene, the freezing point decreased by 0.41 degrees C.If the K_{fp} of the solvent is 5.12 K/m, calculate the molar mass of the unknown solute.

a. 20.23 g/mol

b. 202.3 g/mol

c. 2023 g/mol

d. not enough information is given

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