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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![**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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5014d6a8-882c-467c-b1c9-5ade3eaa5554%2F9f3668c1-3b3f-4279-8f35-56d30db95e8b%2F65wo4m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**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.
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