Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding182g of glycerin to 338mL of H2O at 40.0°C? The vapor pressure of pure water at 40.0°C is 55.32 torr and its density is 0.992 g/cm3

To find the vapor pressure of the solution, we can use Raoult's Law, which states that the vapor pressure of the solution (\(P_{solution}\)) is equal to the vapor pressure of the pure solvent (\(P^0_{solvent}\)) multiplied by the mole fraction of the solvent (\(X_{solvent}\)):

\[
P_{solution} = X_{solvent} \times P^0_{solvent}
\]

### Step 1: Calculate the moles of glycerin and water

1. **Moles of glycerin (C\(_3\)H\(_8\)O\(_3\))**:
   - Molar mass of glycerin \(= 3 \times 12.01 + 8 \times 1.01 + 3 \times 16.00 = 92.09 \, \text{g/mol}\)
   - Moles of glycerin \(n_{glycerin}\):

   \[
   n_{glycerin} = \frac{182 \, \text{g}}{92.09 \, \text{g/mol}} \approx 1.976 \, \text{mol}
   \]

2. **Moles of water (H\(_2\)O)**:
   - First, find the mass of water using its density:
     \[
     \text{Density of water} = 0.992 \, \text{g/cm}^3 = 0.992 \, \text{g/mL}
     \]
     \[
     \text{Mass of water} = 338 \, \text{mL} \times 0.992 \, \text{g/mL} = 335.296 \, \text{g}
     \]
   - Molar mass of water \(= 2 \times 1.01 + 16.00 = 18.02 \, \text{g/mol}\)
   - Moles of water \(n_{water}\):
     \[
     n_{water} = \frac{335.296 \, \text{g}}{18.02 \, \text{g/mol}} \approx 18.608 \, \text{mol}
     \]

### Step 2: Calculate the mole fraction of water

\[
X_{water} = \frac{n_{water}}{n_{water} + n_{glycerin}} = \frac{18.608}{18.608 + 1.976} \approx \frac{18.608}{20.584} \approx 0.904
\]

### Step 3: Apply Raoult's Law

\[
P_{solution} = X_{water} \times P^0_{water} = 0.904 \times 55.32 \, \text{torr} \approx 50.01 \, \text{torr}
\]

### Final Answer:
The vapor pressure of the solution is approximately **50.01 torr** at 40.0°C.

Would you like more details or have any questions about this process?

Here are some related questions you might find interesting:

1. How would the vapor pressure change if the temperature was increased?
2. What happens to the vapor pressure if a volatile solute is used instead of glycerin?
3. How does the addition of more glycerin affect the vapor pressure?
4. What is the significance of glycerin being nonvolatile in this calculation?
5. How would you calculate the boiling point elevation for this solution?

**Tip:** In problems involving Raoult's Law, always verify whether the solute is volatile or nonvolatile, as this impacts the calculation of the vapor pressure.

Learn how to calculate the vapor pressure of a solution containing 182g of glycerin in 338mL of water at 40.0°C using Raoult's Law. The vapor pressure of pure water at 40.0°C is 55.32 torr, and the density of water is 0.992 g/cm³.

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