The vapor pressure of ethanol is 54.68 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in ethanol is saccharin Calculate the vapor pressure of the solution at 25 °C when 13.39 grams of saccharin, C-H5NO3S (183.2 g/mol), are dissolved in 204.9 grams of ethanol. ethanol = CH3CH2OH = 46.07 g/mol. VP(solution) = mm Hg

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### Calculating Vapor Pressure of a Solution

**Problem Statement:**
The vapor pressure of **ethanol** is 54.68 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in ethanol is **saccharin**. 

Calculate the vapor pressure of the solution at 25 °C when 13.39 grams of saccharin, \( C_7H_5NO_3S \) (183.2 g/mol), are dissolved in 204.9 grams of ethanol.

**Given:**
- Ethanol = \( C_2H_5OH \) = 46.07 g/mol
- Vapor Pressure of ethanol = 54.68 mm Hg
- Mass of saccharin = 13.39 g
- Mass of ethanol = 204.9 g

**Solution:**
To find the vapor pressure of the solution, use Raoult's Law:

\[
\text{VP(solution)} = \text{VP(pure solvent)} \times \text{mole fraction of solvent}
\]

### Steps:

1. **Calculate Moles:**
   - Moles of saccharin: 
     \[
     \left(\frac{13.39 \text{ g}}{183.2 \text{ g/mol}}\right) = 0.0731 \text{ moles}
     \]
   - Moles of ethanol: 
     \[
     \left(\frac{204.9 \text{ g}}{46.07 \text{ g/mol}}\right) = 4.448 \text{ moles}
     \]

2. **Calculate Mole Fraction of Ethanol:**
   - Total moles = 0.0731 + 4.448 = 4.521 moles
   - Mole fraction of ethanol:
     \[
     \left(\frac{4.448}{4.521}\right) = 0.9839
     \]

3. **Calculate Vapor Pressure of the Solution:**
   - Using Raoult's Law:
     \[
     \text{VP(solution)} = 54.68 \times 0.9839 = 53.80 \text{ mm Hg}
     \]

**Conclusion:**
The vapor pressure of the solution containing saccharin and ethanol at 25 °C is 53.80 mm Hg.

**Interactive Components:
Transcribed Image Text:### Calculating Vapor Pressure of a Solution **Problem Statement:** The vapor pressure of **ethanol** is 54.68 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in ethanol is **saccharin**. Calculate the vapor pressure of the solution at 25 °C when 13.39 grams of saccharin, \( C_7H_5NO_3S \) (183.2 g/mol), are dissolved in 204.9 grams of ethanol. **Given:** - Ethanol = \( C_2H_5OH \) = 46.07 g/mol - Vapor Pressure of ethanol = 54.68 mm Hg - Mass of saccharin = 13.39 g - Mass of ethanol = 204.9 g **Solution:** To find the vapor pressure of the solution, use Raoult's Law: \[ \text{VP(solution)} = \text{VP(pure solvent)} \times \text{mole fraction of solvent} \] ### Steps: 1. **Calculate Moles:** - Moles of saccharin: \[ \left(\frac{13.39 \text{ g}}{183.2 \text{ g/mol}}\right) = 0.0731 \text{ moles} \] - Moles of ethanol: \[ \left(\frac{204.9 \text{ g}}{46.07 \text{ g/mol}}\right) = 4.448 \text{ moles} \] 2. **Calculate Mole Fraction of Ethanol:** - Total moles = 0.0731 + 4.448 = 4.521 moles - Mole fraction of ethanol: \[ \left(\frac{4.448}{4.521}\right) = 0.9839 \] 3. **Calculate Vapor Pressure of the Solution:** - Using Raoult's Law: \[ \text{VP(solution)} = 54.68 \times 0.9839 = 53.80 \text{ mm Hg} \] **Conclusion:** The vapor pressure of the solution containing saccharin and ethanol at 25 °C is 53.80 mm Hg. **Interactive Components:
**Problem Statement:**

The vapor pressure of ethanol is 54.68 mm Hg at 25°C.

**Question:**

How many grams of estrogen (estradiol), C₁₈H₂₄O₂, a nonvolatile, nonelectrolyte (MW = 272.4 g/mol), must be added to 245.3 grams of ethanol to reduce the vapor pressure to 53.74 mm Hg?

**Given Data:**

- Ethanol formula: CH₃CH₂OH
- Molar mass of ethanol: 46.07 g/mol

**Calculation Required:**

\[\_\_\_\_\_\_\_\_\_\] g of estrogen

**Options for Submission:**

- Submit Answer
- Retry Entire Group
- 9 more group attempts remaining

**Technical Support:**

Cengage Learning | Cengage Technical Support

(Note: Below the problem, navigation options such as "Previous" and "Next" are available for moving through the questions.)
Transcribed Image Text:**Problem Statement:** The vapor pressure of ethanol is 54.68 mm Hg at 25°C. **Question:** How many grams of estrogen (estradiol), C₁₈H₂₄O₂, a nonvolatile, nonelectrolyte (MW = 272.4 g/mol), must be added to 245.3 grams of ethanol to reduce the vapor pressure to 53.74 mm Hg? **Given Data:** - Ethanol formula: CH₃CH₂OH - Molar mass of ethanol: 46.07 g/mol **Calculation Required:** \[\_\_\_\_\_\_\_\_\_\] g of estrogen **Options for Submission:** - Submit Answer - Retry Entire Group - 9 more group attempts remaining **Technical Support:** Cengage Learning | Cengage Technical Support (Note: Below the problem, navigation options such as "Previous" and "Next" are available for moving through the questions.)
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