A 0.500 gram sample was dissolved in 50.0 grams of benzene and the resulting solution froze at 5.06 ° C. What is the molar mass of the sample if pure benzene freezes at 5.48 ° C?
K_f(benzene) = 5.12 ° C/ m .

Question

Quizplus · Accepted Answer

The freezing point depression ($\Delta T_f$) is calculated using the formula $\Delta T_f = K_f \times m$, where $K_f$ is the cryoscopic constant (5.12 °C/m for benzene) and $m$ is the molality of the solution. The change in freezing point is the difference between the freezing point of pure benzene (5.48 °C) and the freezing point of the solution (5.06 °C), which is $5.48 °C - 5.06 °C = 0.42 °C$. To find the molality, we rearrange the formula to $m = \Delta T_f / K_f = 0.42 °C / 5.12 °C/m = 0.08203 m$. Molality is moles of solute per kilogram of solvent, so $0.08203 m = \frac{moles\ of\ solute}{0.050\ kg}$. Solving for moles of solute gives $0.08203 m \times 0.050\ kg = 0.0041015\ moles$. The molar mass is then the mass of the sample divided by the moles of solute, $0.500\ g / 0.0041015\ moles = 121.9\ g/mol$, which rounds to 122 g/mol.

A 0500 Gram Sample Was Dissolved in 50