**Freezing Point Depression and Molar Mass Calculation**When 2.00 grams of an unknown compound is mixed with 75.00 grams of benzene, the freezing point of the solution drops from 5.53°C to 4.90°C. To determine the molar mass of the compound, we can use the equation for freezing point depression:\[\Delta T_f = -K_f \times m\]Where:- \(\Delta T_f\) is the change in freezing point.- \(K_f\) is the freezing point depression constant for benzene, given as 5.12°C kg/mol. - \(m\) is the molality of the solution.**Steps to Calculate the Molar Mass:**1. **Calculate \(\Delta T_f\):** \[ \Delta T_f = 5.53°C - 4.90°C = 0.63°C \]2. **Determine the molality (\(m\)):** Use the formula: \[ m = \frac{\Delta T_f}{K_f} = \frac{0.63°C}{5.12°C \, \text{kg/mol}} \]3. **Calculate moles of solute:** The molality (\(m\)) is moles of solute per kg of solvent. Use the mass of benzene (75.00 g = 0.075 kg) to find the moles of solute.4. **Determine the molar mass of the compound:** Use the formula: \[ \text{Molar mass} = \frac{\text{mass of solute (in grams)}}{\text{moles of solute}} \]This process will yield the molar mass of the unknown compound.

2.00 g of an unknown compound is mixed to form a homogeneous solution with 75.00 g of benzene. This reduces the freezing point from 5.53 to 4.90 °C. What is the molar mass of the compound? AT;= – Kfx m; Kj= 5.12 for benzene

2.00 g of an unknown compound is mixed to form a homogeneous solution with 75.00 g of benzene. This reduces the freezing point from 5.53 to 4.90 °C. What is the molar mass of the compound? AT;= – Kfx m; Kj= 5.12 for benzene

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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