Multiple choice

A solution is prepared by mixing ${\text{8}}{\text{.5}}\,{\text{g}}$ of ${\text{C}}{{\text{H}}{\text{2}}}{\text{C}}{{\text{l}}_2}$ and $11.95\,{\text{g}}$ of ${\text{CHC}}{{\text{l}}{\text{3}}}$. If vapour pressure of ${\text{C}}{{\text{H}}{\text{2}}}{\text{C}}{{\text{l}}{\text{2}}}$ and ${\text{CHC}}{{\text{l}}3}$ at ${\text{298}}\,{\text{K}}$ are $415$ and $200\,{\text{mm}}$ Hg respecetively, then mole fraction of ${\text{CHC}}{{\text{l}}{\text{3}}}$ in vapour form is:

  1. $0.162$
  2. $0.675$
  3. $0.325$
  4. $0.486$
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C Correct answer
Explanation

The two substances have approximately equal mole amounts, so their liquid-phase mole fractions are both about 0.5. Their partial pressures are therefore about 207.5 mm Hg and 100 mm Hg, giving a chloroform vapor mole fraction of 100/(207.5 + 100) = 0.325.

AI explanation

Calculate the moles of dichloromethane by dividing 8.5 g by its molar mass of 85 g/mol to get 0.1 mol, and the moles of chloroform by dividing 11.95 g by its molar mass of 119.5 g/mol to get 0.1 mol. The mole fraction of chloroform in the liquid mixture is 0.1 divided by the total 0.2 moles, which equals 0.5. Using the modified Raoult's law for the vapor phase, multiply the liquid mole fraction of 0.5 by the vapor pressure of 200 mm Hg to get a partial pressure of 100 mm Hg. The total vapor pressure is the sum of the partial pressures (0.5 times 415 plus 100), which equals 307.5 mm Hg, making the mole fraction of chloroform in the vapor phase 100 divided by 307.5, or 0.325.

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