Solving for Vapor Pressure of a Solution Using Raoult's Law

This problem involves calculating the vapor pressure of a solution containing a non-volatile solute (urea) dissolved in a solvent (water). We can solve this using Raoult's Law, which describes the relationship between the mole fraction of the solvent and the vapor pressure of the solution.

Problem Analysis

We are given:

• The solute is urea ($CO(NH_2)_2$), which is non-volatile.
• The mass of urea is $18 \ g$.
• The molar mass of urea is $60 \ g/mol$.
• The solvent is water ($H_2O$).
• The mass of water is $180 \ g$.
• The molar mass of water is $18 \ g/mol$.
• The vapor pressure of pure water ($P^0_{water}$) is $50 \ mmHg$.
• We need to find the vapor pressure of the solution ($P_{solution}$).

Key Concepts and Formulas

The core principle here is Raoult's Law for solutions with non-volatile solutes. It states that the vapor pressure of the solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.

The formula is:

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

Where:

• $P_{solution}$ is the vapor pressure of the solution.
• $X_{solvent}$ is the mole fraction of the solvent (water in this case).
• $P^0_{solvent}$ is the vapor pressure of the pure solvent (pure water).

To find the mole fraction of the solvent ($X_{solvent}$), we need the moles of the solvent ($n_{solvent}$) and the moles of the solute ($n_{solute}$):

$X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}}$

The number of moles ($n$) is calculated as:

$n = \frac{\text{mass}}{\text{molar mass}}$

Step-by-Step Calculations

1. Calculate Moles of Solute (Urea)

Using the given mass and molar mass of urea:

$n_{urea} = \frac{18 \ g}{60 \ g/mol} = 0.3 \ mol$

2. Calculate Moles of Solvent (Water)

Using the given mass and molar mass of water:

$n_{water} = \frac{180 \ g}{18 \ g/mol} = 10 \ mol$

3. Calculate Total Moles

The total number of moles in the solution is the sum of the moles of water and urea:

$n_{total} = n_{water} + n_{urea} = 10 \ mol + 0.3 \ mol = 10.3 \ mol$

4. Calculate Mole Fraction of Solvent (Water)

Now, we find the mole fraction of water:

$X_{water} = \frac{n_{water}}{n_{total}} = \frac{10 \ mol}{10.3 \ mol}$

$X_{water} \approx 0.97087$

5. Calculate Vapor Pressure of the Solution

Apply Raoult's Law using the calculated mole fraction of water and the given vapor pressure of pure water:

$P_{solution} = X_{water} \times P^0_{water}$

$P_{solution} = \frac{10}{10.3} \times 50 \ mmHg$

$P_{solution} \approx 0.97087 \times 50 \ mmHg$

$P_{solution} \approx 48.5435 \ mmHg$

Conclusion

Rounding the result to two decimal places, the vapor pressure of the solution is approximately $48.54 \ mmHg$. This value is lower than the vapor pressure of pure water, as expected when a non-volatile solute is added.

Detailed breakdown of the calculation:

Total Moles = Moles of Water + Moles of Urea = $10 \ mol + 0.3 \ mol = 10.3 \ mol$

Mole Fraction of Water ($X_{water}$) = $\frac{\text{Moles of Water}}{\text{Total Moles}} = \frac{10}{10.3}$

Vapor Pressure of Solution ($P_{solution}$) = $X_{water} \times P^0_{water}$ = $\frac{10}{10.3} \times 50 \ mmHg \approx 48.54 \ mmHg$