Answer :
To solve the problem of finding the vapor pressure of the solution containing benzene and camphor, we can use Raoult's Law. Raoult's Law states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.
Here is a step-by-step approach:
1. Gather Given Information:
- Vapor pressure of pure benzene at [tex]\(26.1^\circ C = 100.0 \, \text{mmHg}\)[/tex].
- Mass of camphor ([tex]\(C_{10}H_{16}O\)[/tex]) = 24.6 g.
- Mass of benzene ([tex]\(C_6H_6\)[/tex]) = 98.5 g.
- Molar mass of camphor = 152.23 g/mol.
- Molar mass of benzene = 78.11 g/mol.
2. Calculate the Moles of each Substance:
- Moles of camphor:
[tex]\[
\text{Moles of camphor} = \frac{\text{mass of camphor}}{\text{molar mass of camphor}} = \frac{24.6 \, \text{g}}{152.23 \, \text{g/mol}} \approx 0.162 \, \text{mol}
\][/tex]
- Moles of benzene:
[tex]\[
\text{Moles of benzene} = \frac{\text{mass of benzene}}{\text{molar mass of benzene}} = \frac{98.5 \, \text{g}}{78.11 \, \text{g/mol}} \approx 1.261 \, \text{mol}
\][/tex]
3. Calculate the Total Moles in the Solution:
[tex]\[
\text{Total moles} = \text{moles of camphor} + \text{moles of benzene} = 0.162 \, \text{mol} + 1.261 \, \text{mol} \approx 1.423 \, \text{mol}
\][/tex]
4. Determine the Mole Fraction of Benzene:
[tex]\[
\text{Mole fraction of benzene} = \frac{\text{moles of benzene}}{\text{total moles}} = \frac{1.261 \, \text{mol}}{1.423 \, \text{mol}} \approx 0.886
\][/tex]
5. Calculate the Vapor Pressure of the Solution using Raoult’s Law:
[tex]\[
\text{Vapor pressure of the solution} = \text{vapor pressure of pure benzene} \times \text{mole fraction of benzene}
\][/tex]
[tex]\[
\text{Vapor pressure of the solution} = 100.0 \, \text{mmHg} \times 0.886 \approx 88.64 \, \text{mmHg}
\][/tex]
Thus, the vapor pressure of the benzene-camphor solution is approximately 88.64 mmHg.
Here is a step-by-step approach:
1. Gather Given Information:
- Vapor pressure of pure benzene at [tex]\(26.1^\circ C = 100.0 \, \text{mmHg}\)[/tex].
- Mass of camphor ([tex]\(C_{10}H_{16}O\)[/tex]) = 24.6 g.
- Mass of benzene ([tex]\(C_6H_6\)[/tex]) = 98.5 g.
- Molar mass of camphor = 152.23 g/mol.
- Molar mass of benzene = 78.11 g/mol.
2. Calculate the Moles of each Substance:
- Moles of camphor:
[tex]\[
\text{Moles of camphor} = \frac{\text{mass of camphor}}{\text{molar mass of camphor}} = \frac{24.6 \, \text{g}}{152.23 \, \text{g/mol}} \approx 0.162 \, \text{mol}
\][/tex]
- Moles of benzene:
[tex]\[
\text{Moles of benzene} = \frac{\text{mass of benzene}}{\text{molar mass of benzene}} = \frac{98.5 \, \text{g}}{78.11 \, \text{g/mol}} \approx 1.261 \, \text{mol}
\][/tex]
3. Calculate the Total Moles in the Solution:
[tex]\[
\text{Total moles} = \text{moles of camphor} + \text{moles of benzene} = 0.162 \, \text{mol} + 1.261 \, \text{mol} \approx 1.423 \, \text{mol}
\][/tex]
4. Determine the Mole Fraction of Benzene:
[tex]\[
\text{Mole fraction of benzene} = \frac{\text{moles of benzene}}{\text{total moles}} = \frac{1.261 \, \text{mol}}{1.423 \, \text{mol}} \approx 0.886
\][/tex]
5. Calculate the Vapor Pressure of the Solution using Raoult’s Law:
[tex]\[
\text{Vapor pressure of the solution} = \text{vapor pressure of pure benzene} \times \text{mole fraction of benzene}
\][/tex]
[tex]\[
\text{Vapor pressure of the solution} = 100.0 \, \text{mmHg} \times 0.886 \approx 88.64 \, \text{mmHg}
\][/tex]
Thus, the vapor pressure of the benzene-camphor solution is approximately 88.64 mmHg.
