** 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:

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

**Step 1:**

ΔT_{f} = **3.66˚C** i = **1** *(benzene is a non-electrolyte)*

**m = ??** K_{f} = **1.86 ˚C • kg/mol** *(can be found in textbooks or online)*

Solving for **molality**:

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

$\mathbf{3}\mathbf{.}\mathbf{66}\mathbf{\xb0}\mathbf{C}\mathbf{=}\left(\mathbf{1}\right)\left(\mathbf{m}\right)\left(\mathbf{5}\mathbf{.}\mathbf{12}\mathbf{}\frac{\mathbf{\xb0}\mathbf{C}}{\mathbf{m}}\right)$

$\mathbf{m}\mathbf{=}\frac{\mathbf{3}\mathbf{.}\mathbf{66}\mathbf{}\overline{)\mathbf{\xb0}\mathbf{C}}}{\left(\mathbf{5}\mathbf{.}\mathbf{12}\mathbf{}{\displaystyle \frac{\overline{)\mathbf{\xb0}\mathbf{C}}}{\mathbf{m}}}\right)}$** = 0.7148 mol compound**_{ }/ kg benzene

**Step 3:**

A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by 3.66 °C. Calculate the molar mass of the compound.

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