The freezing point depression constant for benzene is 5.06 and the freezing point gni sdoig of pure benzene is 5.51°C. If a solution containing 2.241 g of a solid dissolved into 50.0 g of benzene has a freezing point of 4.88°C, what is the molar mass of the solid?

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**Title: Freezing Point Depression**

**Problem 4:**

The freezing point depression constant for benzene is \(5.06 \, \frac{°C}{m}\), and the freezing point of pure benzene is \(5.51°C\). If a solution containing \(2.241 \, \text{g}\) of a solid dissolved into \(50.0 \, \text{g}\) of benzene has a freezing point of \(4.88°C\), what is the molar mass of the solid?

---

This problem deals with the concept of freezing point depression, which is the decrease in the freezing point of a solvent when a solute is dissolved in it. The freezing point depression constant, often denoted as \(K_f\), is specific to each solvent and relates to how much the freezing point will lower for each molal unit of solute added.

To solve this problem, you can use the formula:

\[
\Delta T_f = K_f \cdot m
\]

Where:
- \(\Delta T_f\) is the change in freezing point (\(5.51°C - 4.88°C\)),
- \(K_f\) is the freezing point depression constant,
- \(m\) is the molality of the solution.

The molality (\(m\)) is calculated as the moles of solute per kilogram of solvent. Finally, using the definition of molality and the given information, you can determine the molar mass of the solute in the solution.
Transcribed Image Text:**Title: Freezing Point Depression** **Problem 4:** The freezing point depression constant for benzene is \(5.06 \, \frac{°C}{m}\), and the freezing point of pure benzene is \(5.51°C\). If a solution containing \(2.241 \, \text{g}\) of a solid dissolved into \(50.0 \, \text{g}\) of benzene has a freezing point of \(4.88°C\), what is the molar mass of the solid? --- This problem deals with the concept of freezing point depression, which is the decrease in the freezing point of a solvent when a solute is dissolved in it. The freezing point depression constant, often denoted as \(K_f\), is specific to each solvent and relates to how much the freezing point will lower for each molal unit of solute added. To solve this problem, you can use the formula: \[ \Delta T_f = K_f \cdot m \] Where: - \(\Delta T_f\) is the change in freezing point (\(5.51°C - 4.88°C\)), - \(K_f\) is the freezing point depression constant, - \(m\) is the molality of the solution. The molality (\(m\)) is calculated as the moles of solute per kilogram of solvent. Finally, using the definition of molality and the given information, you can determine the molar mass of the solute in the solution.
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