A student prepares a solution by dissolving 2.09 g of para-dichlorobenzene (CçH_C,, M = 147.0 g/mol) in 50.0 mL of cyclohexane. What is the freezing point (in °C) of the prepared solution? Pure cyclohexane has a freezing point of 6.47°C, a freezing point depression constant of 20.0°C/m, and a density of 1.25 g/mL. Freezing point PC %3D
A student prepares a solution by dissolving 2.09 g of para-dichlorobenzene (CçH_C,, M = 147.0 g/mol) in 50.0 mL of cyclohexane. What is the freezing point (in °C) of the prepared solution? Pure cyclohexane has a freezing point of 6.47°C, a freezing point depression constant of 20.0°C/m, and a density of 1.25 g/mL. Freezing point PC %3D
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![### Freezing Point Depression Calculation
A student prepares a solution by dissolving 2.09 g of para-dichlorobenzene (C₆H₄Cl₂, \( M = 147.0 \, \text{g/mol} \)) in 50.0 mL of cyclohexane. What is the freezing point (in °C) of the prepared solution? Pure cyclohexane has a freezing point of 6.47°C, a freezing point depression constant of 20.0°C/m, and a density of 1.25 g/mL.
#### Calculation Steps
1. **Determine the molality (m) of the solution:**
- First, calculate the moles of para-dichlorobenzene:
\[
\text{Moles of para-dichlorobenzene} = \frac{2.09 \, \text{g}}{147.0 \, \text{g/mol}} = 0.01423 \, \text{mol}
\]
- Convert the volume of cyclohexane from mL to kg:
\[
50.0 \, \text{mL} \times 1.25 \, \text{g/mL} = 62.5 \, \text{g} = 0.0625 \, \text{kg}
\]
- Calculate the molality (m):
\[
\text{molality} = \frac{0.01423 \, \text{mol}}{0.0625 \, \text{kg}} = 0.22768 \, \text{m}
\]
2. **Apply the freezing point depression formula:**
\[
\Delta T_f = K_f \times m
\]
Where:
- \( \Delta T_f \) is the freezing point depression.
- \( K_f \) is the freezing point depression constant (20.0°C/m).
- \( m \) is the molality of the solution calculated above.
\[
\Delta T_f = 20.0 \, \frac{°C}{m} \times 0.22768 \, m = 4.5536 \, °C
\]
3. **Determine the new freezing point of the solution:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7eaef044-20a7-420f-b57c-5d0c5f8d0743%2F6a36ea10-9aa4-45c8-b4c5-71317acb8852%2F4b8ce6e_reoriented.jpeg&w=3840&q=75)
Transcribed Image Text:### Freezing Point Depression Calculation
A student prepares a solution by dissolving 2.09 g of para-dichlorobenzene (C₆H₄Cl₂, \( M = 147.0 \, \text{g/mol} \)) in 50.0 mL of cyclohexane. What is the freezing point (in °C) of the prepared solution? Pure cyclohexane has a freezing point of 6.47°C, a freezing point depression constant of 20.0°C/m, and a density of 1.25 g/mL.
#### Calculation Steps
1. **Determine the molality (m) of the solution:**
- First, calculate the moles of para-dichlorobenzene:
\[
\text{Moles of para-dichlorobenzene} = \frac{2.09 \, \text{g}}{147.0 \, \text{g/mol}} = 0.01423 \, \text{mol}
\]
- Convert the volume of cyclohexane from mL to kg:
\[
50.0 \, \text{mL} \times 1.25 \, \text{g/mL} = 62.5 \, \text{g} = 0.0625 \, \text{kg}
\]
- Calculate the molality (m):
\[
\text{molality} = \frac{0.01423 \, \text{mol}}{0.0625 \, \text{kg}} = 0.22768 \, \text{m}
\]
2. **Apply the freezing point depression formula:**
\[
\Delta T_f = K_f \times m
\]
Where:
- \( \Delta T_f \) is the freezing point depression.
- \( K_f \) is the freezing point depression constant (20.0°C/m).
- \( m \) is the molality of the solution calculated above.
\[
\Delta T_f = 20.0 \, \frac{°C}{m} \times 0.22768 \, m = 4.5536 \, °C
\]
3. **Determine the new freezing point of the solution:**
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