Vapour pressure of $CHCl_3$ and $CH_2Cl_2$ at 298 K are 200 mmHg and 415 mmHg respectively. Calculate the vapour pressure of the solution prepared by mixing 25.5 g $CHCl_3$ and 40 g $CH_2Cl_2$ at 298 K.

347.9 mmHg

The correct answer is Option (4) → 347.9 mmHg

To calculate the vapor pressure of a solution prepared by mixing CHCl₃ (chloroform) and CH₂Cl₂ (dichloromethane), we can use Raoult's Law, which states:

\(P_{\text{solution}} = \chi _{\text{solvent}} \cdot P^0_{\text{solvent}} + \chi _{\text{solute}} \cdot P^0_{\text{solute}}\)

Where:

\( P_{\text{solution}} \) = vapor pressure of the solution

\( \chi \) = mole fraction of the component in the solution

\( P^0 \) = vapor pressure of the pure component

Molar mass of CHCl₃ = \( 1 \times 12 + 1 \times 1 + 3 \times 35.5 = 12 + 1 + 106.5 = 119.5 \, \text{g/mol} \)

Molar mass of CH₂Cl₂ = \( 1 \times 12 + 2 \times 1 + 2 \times 35.5 = 12 + 2 + 71 = 85 \, \text{g/mol} \)

Moles of CHCl₃:

\(\text{Moles of CHCl}_3 = \frac{25.5 \, \text{g}}{119.5 \, \text{g/mol}} \approx 0.213 \, \text{mol}\)

Moles of CH₂Cl₂:

\(\text{Moles of CH}_2\text{Cl}_2 = \frac{40 \, \text{g}}{85 \, \text{g/mol}} \approx 0.471 \, \text{mol}\)

\(\text{Total moles} = \text{Moles of CHCl}_3 + \text{Moles of CH}_2\text{Cl}_2 = 0.213 + 0.471 = 0.684 \, \text{mol}\)

Mole fraction of CHCl₃ (\(\chi _{\text{CHCl}_3}\)):
\(\chi _{\text{CHCl}_3} = \frac{0.213}{0.684} \approx 0.311\)

Mole fraction of CH₂Cl₂ (\(\chi _{\text{CH}_2\text{Cl}_2}\)):

\(\chi _{\text{CH}_2\text{Cl}_2} = \frac{0.471}{0.684} \approx 0.689\)

Given:

\(P^0_{\text{CHCl}_3} = 200 \, \text{mmHg}\)

\(P^0_{\text{CH}_2\text{Cl}_2} = 415 \, \text{mmHg}\)

Now plug in the values:

\(P_{\text{solution}} = (\chi _{\text{CHCl}_3} \cdot P^0_{\text{CHCl}_3}) + (\chi _{\text{CH}_2\text{Cl}_2} \cdot P^0_{\text{CH}_2\text{Cl}_2})\)

For CHCl₃:

\(P_{\text{CHCl}_3} = 0.311 \times 200 \approx 62.2 \, \text{mmHg}\)

For CH₂Cl₂:

\(P_{\text{CH}_2\text{Cl}_2} = 0.689 \times 415 \approx 286.9 \, \text{mmHg}\)

Final Vapor Pressure Calculation:

\(P_{\text{solution}} = 62.2 + 286.9 \approx 349.1 \, \text{mmHg}\)

Conclusion

The closest answer choice based on the calculations would be 347.9 mmHg.

Thus, the vapor pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K is approximately 347.9 mmH