When 1.504 g of a solute is dissolved in 5.012 g of solvent (benzene), the solution freezes at -7.05 °C. The pure solvent has a freezing point of 5.50°C. The Ki for the solvent is 5.12 °C/m. What is the molar mass of the solute?
When 1.504 g of a solute is dissolved in 5.012 g of solvent (benzene), the solution freezes at -7.05 °C. The pure solvent has a freezing point of 5.50°C. The Ki for the solvent is 5.12 °C/m. What is the molar mass of the solute?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![When 1.504 g of a solute is dissolved in 5.012 g of solvent (benzene), the solution freezes at -7.05 °C. The pure solvent has a freezing point of 5.50°C. The Kf for the solvent is 5.12 °C/m. What is the molar mass of the solute?
### Explanation
This problem deals with the concept of freezing point depression, which is a colligative property. It occurs when a solute is added to a solvent, lowering the freezing point of the resulting solution compared to the pure solvent.
#### Given Data:
- Mass of solute = 1.504 g
- Mass of solvent (benzene) = 5.012 g
- Freezing point of the solution = -7.05 °C
- Freezing point of pure solvent = 5.50 °C
- Freezing point depression constant (Kf) for benzene = 5.12 °C/m
#### Calculation Steps:
1. **Calculate the Freezing Point Depression (ΔTf):**
\[
ΔTf = \text{Freezing point of pure solvent} - \text{Freezing point of solution}
\]
\[
ΔTf = 5.50 °C - (-7.05 °C) = 12.55 °C
\]
2. **Apply the Formula for Freezing Point Depression:**
\[
ΔTf = Kf \times m
\]
Where \( m \) is the molality of the solution.
3. **Calculate the Molality (m):**
\[
m = \frac{ΔTf}{Kf} = \frac{12.55 °C}{5.12 °C/m} = 2.45 \, \text{mol/kg}
\]
4. **Calculate the Moles of Solute:**
\[
\text{Molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}}
\]
\[
2.45 \, \text{mol/kg} = \frac{\text{moles of solute}}{0.005012 \, \text{kg}}
\]
\[
\text{Moles of solute} = 2.45 \, \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb9531bb1-7383-4fb6-bd2d-66ab886f3f55%2F438535aa-a59f-4d38-bd27-3dfe162d4fc6%2Fw8dfflo_processed.png&w=3840&q=75)
Transcribed Image Text:When 1.504 g of a solute is dissolved in 5.012 g of solvent (benzene), the solution freezes at -7.05 °C. The pure solvent has a freezing point of 5.50°C. The Kf for the solvent is 5.12 °C/m. What is the molar mass of the solute?
### Explanation
This problem deals with the concept of freezing point depression, which is a colligative property. It occurs when a solute is added to a solvent, lowering the freezing point of the resulting solution compared to the pure solvent.
#### Given Data:
- Mass of solute = 1.504 g
- Mass of solvent (benzene) = 5.012 g
- Freezing point of the solution = -7.05 °C
- Freezing point of pure solvent = 5.50 °C
- Freezing point depression constant (Kf) for benzene = 5.12 °C/m
#### Calculation Steps:
1. **Calculate the Freezing Point Depression (ΔTf):**
\[
ΔTf = \text{Freezing point of pure solvent} - \text{Freezing point of solution}
\]
\[
ΔTf = 5.50 °C - (-7.05 °C) = 12.55 °C
\]
2. **Apply the Formula for Freezing Point Depression:**
\[
ΔTf = Kf \times m
\]
Where \( m \) is the molality of the solution.
3. **Calculate the Molality (m):**
\[
m = \frac{ΔTf}{Kf} = \frac{12.55 °C}{5.12 °C/m} = 2.45 \, \text{mol/kg}
\]
4. **Calculate the Moles of Solute:**
\[
\text{Molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}}
\]
\[
2.45 \, \text{mol/kg} = \frac{\text{moles of solute}}{0.005012 \, \text{kg}}
\]
\[
\text{Moles of solute} = 2.45 \, \
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