The vapour pressure os chloroform \((CHCl_3)\) and dichloromethane \(CH_2Cl_2\) at 298 K are 400 mm Hg and 615 mm Hg, respectively. Ehat is the mole fraction of \(CHCl_3\) in vapour phase in solution prepared by mixing 30 g of \(CHCl_3\) and 45 g of \(CH_2Cl_2\)?
[Consider the mass of the given atoms as: C =12.01u, H = 1.01 u and Cl = 35.45 u]
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 0.23.
From the given data,
Molar mass of \(CHCl_3\):
\(\text{C} = 12.01 \, \text{g/mol}, \, \text{H} = 1.01 \, \text{g/mol}, \, \text{Cl} = 35.45 \, \text{g/mol}\)
\(\text{Molar mass of } CHCl_3 = 12.01 + 1.01 + (3 \times 35.45) = 119.38 \, \text{g/mol}\)
Molar mass of \(CH_2Cl_2\):
\(\text{C} = 12.01 \, \text{g/mol}, \, \text{H} = 1.01 \, \text{g/mol}, \, \text{Cl} = 35.45 \, \text{g/mol}\)
\(\text{Molar mass of } CH_2Cl_2 = 12.01 + (2 \times 1.01) + (2 \times 35.45) = 84.93 \, \text{g/mol}\)
Number of moles of \(CHCl_3\):
\(n_{CHCl_3} = \frac{30 \, \text{g}}{119.38 \, \text{g/mol}} = 0.251 \, \text{mol}\)
Number of moles of \(CH_2Cl_2\):
\(n_{CH_2Cl_2} = \frac{45 \, \text{g}}{84.93 \, \text{g/mol}} = 0.53 \, \text{mol}\)
\(\text{Total moles } n_{total} = 0.251 + 0.53 = 0.781 \, \text{mol}\)
\(\text{Mole fraction of } CHCl_3 \, (\chi _{CHCl_3}) = \frac{0.251}{0.781} = 0.321\)
\(\text{Mole fraction of } CH_2Cl_2 \, (\chi _{CH_2Cl_2}) = \frac{0.53}{0.781} = 0.679\)
Applying Raoult's law, we have
\(P_{CHCl_3} = \chi _{CHCl_3} \times P^\circ_{CHCl_3} = 0.321 \times 400 \, \text{mm Hg} = 128.4 \, \text{mm Hg}\)
\(P_{CH_2Cl_2} = \chi _{CH_2Cl_2} \times P^\circ_{CH_2Cl_2} = 0.679 \times 615 \, \text{mm Hg} = 417.585 \, \text{mm Hg}\)
\(P_{total} = P_{CHCl_3} + P_{CH_2Cl_2} = 128.4 \, \text{mm Hg} + 417.585 \, \text{mm Hg} = 545.985 \, \text{mm Hg}\)
\(y_{CHCl_3} = \frac{P_{CHCl_3}}{P_{total}} = \frac{128.4 \, \text{mm Hg}}{545.985 \, \text{mm Hg}} \approx 0.235\)
Conclusion: The mole fraction of \(CHCl_3\) in the vapor phase is approximately 0.23.