When 131. g of glycine (C2H5NO2) are dissolved in 1350. g of a certain mystery liquid X, the freezing point of the solution is 2.00 °C less than the freezing point of pure X. Calculate the mass of iron(III) nitrate (Fe(NO,),) that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor i=3.04 for iron(III) nitrate in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 3 significant digits.
When 131. g of glycine (C2H5NO2) are dissolved in 1350. g of a certain mystery liquid X, the freezing point of the solution is 2.00 °C less than the freezing point of pure X. Calculate the mass of iron(III) nitrate (Fe(NO,),) that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor i=3.04 for iron(III) nitrate in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 3 significant digits.
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
Chapter1: Chemical Foundations
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![### Freezing Point Depression Calculation
When 131. g of glycine \((C_2H_5NO_2)\) are dissolved in 1350. g of a certain mystery liquid \(X\), the freezing point of the solution is 2.00 °C less than the freezing point of pure \(X\).
**Question:**
Calculate the mass of iron(III) nitrate \((Fe(NO_3)_3)\) that must be dissolved in the same mass of \(X\) to produce the same depression in freezing point. The van't Hoff factor \(i = 3.04\) for iron(III) nitrate in \(X\).
Be sure your answer has a unit symbol, if necessary, and round your answer to 3 significant digits.
---
### Solution Steps:
1. **Determine the Molal Freezing Point Depression Constant (\(K_f\))**:
Given data for glycine:
- Mass of glycine = 131. g
- Molar mass of glycine (C₂H₅NO₂) ≈ 75.07 g/mol
- Depression in freezing point = 2.00 °C
Calculate the molality (m):
\[
\text{Molality (m)} = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{\frac{131\, \text{g}}{75.07\, \text{g/mol}}}{1.350\, \text{kg}} \approx 1.292 \text{ mol/kg}
\]
Using the freezing point depression equation:
\[
\Delta T_f = K_f \cdot m \implies K_f = \frac{\Delta T_f}{m} = \frac{2.00\, \text{°C}}{1.292\, \text{mol/kg}} \approx 1.548\, \text{°C·kg/mol}
\]
2. **Calculate the Required Mass of \(Fe(NO_3)_3\)**:
- Molar mass of \(Fe(NO_3)_3\) ≈ 241.86 g/mol
- Molality required to produce the same depression: m = \(\frac{2.00 °C}{1.548°C·kg/mol} ≈ 1.292 \text{ mol/kg}\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1067c048-27d4-462b-825c-b686bc07047b%2F357a6ef0-ad32-431d-83b1-53262f52028f%2Fs4wlr5.png&w=3840&q=75)
Transcribed Image Text:### Freezing Point Depression Calculation
When 131. g of glycine \((C_2H_5NO_2)\) are dissolved in 1350. g of a certain mystery liquid \(X\), the freezing point of the solution is 2.00 °C less than the freezing point of pure \(X\).
**Question:**
Calculate the mass of iron(III) nitrate \((Fe(NO_3)_3)\) that must be dissolved in the same mass of \(X\) to produce the same depression in freezing point. The van't Hoff factor \(i = 3.04\) for iron(III) nitrate in \(X\).
Be sure your answer has a unit symbol, if necessary, and round your answer to 3 significant digits.
---
### Solution Steps:
1. **Determine the Molal Freezing Point Depression Constant (\(K_f\))**:
Given data for glycine:
- Mass of glycine = 131. g
- Molar mass of glycine (C₂H₅NO₂) ≈ 75.07 g/mol
- Depression in freezing point = 2.00 °C
Calculate the molality (m):
\[
\text{Molality (m)} = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{\frac{131\, \text{g}}{75.07\, \text{g/mol}}}{1.350\, \text{kg}} \approx 1.292 \text{ mol/kg}
\]
Using the freezing point depression equation:
\[
\Delta T_f = K_f \cdot m \implies K_f = \frac{\Delta T_f}{m} = \frac{2.00\, \text{°C}}{1.292\, \text{mol/kg}} \approx 1.548\, \text{°C·kg/mol}
\]
2. **Calculate the Required Mass of \(Fe(NO_3)_3\)**:
- Molar mass of \(Fe(NO_3)_3\) ≈ 241.86 g/mol
- Molality required to produce the same depression: m = \(\frac{2.00 °C}{1.548°C·kg/mol} ≈ 1.292 \text{ mol/kg}\
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