Vapour pressure of chloroform ${(CHCl_3)}$ and dichloromethane ${(CH2Cl2)}$ at $25^{\circ}C$ are 200 mmHg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of ${CHCl_3}$ and 40 g of ${CH2Cl2}$ at the same temperature will be: (Molecular mass of ${CHCl_3}$ = 119.5 u and molecular mass of ${CH2Cl2 = 85 \, u}$)

Question

Which one of the following is not correct for an ideal solution?

Answer 1

To calculate the vapor pressure of a solution obtained by mixing chloroform (CHCl_3) and dichloromethane (CH_2Cl_2), we will use Raoult's Law. Raoult's Law states that the partial vapor pressure of each component in the solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. The total vapor pressure of the solution is the sum of the partial pressures of each component.

First, we need to calculate the number of moles of each component:
- Moles of CHCl_3: n_{\text{CHCl}_3} = \frac{\text{mass}}{\text{molar mass}} = \frac{25.5}{119.5} \approx 0.213 \, \text{mol}
- Moles of CH_2Cl_2: n_{\text{CH}_2\text{Cl}_2} = \frac{40}{85} \approx 0.471 \, \text{mol}
Next, calculate the mole fractions of each component in the mixture:
- Total moles n_{\text{total}} = 0.213 + 0.471 = 0.684 \, \text{mol}
- Mole fraction of CHCl_3: X_{\text{CHCl}_3} = \frac{0.213}{0.684} \approx 0.311
- Mole fraction of CH_2Cl_2: X_{\text{CH}_2\text{Cl}_2} = \frac{0.471}{0.684} \approx 0.689
Calculate the partial pressures using Raoult's Law:
- Partial pressure of CHCl_3: P_{\text{CHCl}_3} = 200 \times 0.311 = 62.2 \, \text{mm Hg}
- Partial pressure of CH_2Cl_2: P_{\text{CH}_2\text{Cl}_2} = 41.5 \times 0.689 = 28.58 \, \text{mm Hg}
Finally, calculate the total vapor pressure of the solution by summing the partial pressures:
- P_{\text{total}} = P_{\text{CHCl}_3} + P_{\text{CH}_2\text{Cl}_2} = 62.2 + 28.58 = 90.78 \, \text{mm Hg}

There might be a minor rounding difference due to arithmetic, hence the closest option is 90.38 mm Hg.

Answer 2

To calculate the vapor pressure of a solution obtained by mixing chloroform (CHCl_3) and dichloromethane (CH_2Cl_2), we will use Raoult's Law. Raoult's Law states that the partial vapor pressure of each component in the solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. The total vapor pressure of the solution is the sum of the partial pressures of each component.

First, we need to calculate the number of moles of each component:
- Moles of CHCl_3: n_{\text{CHCl}_3} = \frac{\text{mass}}{\text{molar mass}} = \frac{25.5}{119.5} \approx 0.213 \, \text{mol}
- Moles of CH_2Cl_2: n_{\text{CH}_2\text{Cl}_2} = \frac{40}{85} \approx 0.471 \, \text{mol}
Next, calculate the mole fractions of each component in the mixture:
- Total moles n_{\text{total}} = 0.213 + 0.471 = 0.684 \, \text{mol}
- Mole fraction of CHCl_3: X_{\text{CHCl}_3} = \frac{0.213}{0.684} \approx 0.311
- Mole fraction of CH_2Cl_2: X_{\text{CH}_2\text{Cl}_2} = \frac{0.471}{0.684} \approx 0.689
Calculate the partial pressures using Raoult's Law:
- Partial pressure of CHCl_3: P_{\text{CHCl}_3} = 200 \times 0.311 = 62.2 \, \text{mm Hg}
- Partial pressure of CH_2Cl_2: P_{\text{CH}_2\text{Cl}_2} = 41.5 \times 0.689 = 28.58 \, \text{mm Hg}
Finally, calculate the total vapor pressure of the solution by summing the partial pressures:
- P_{\text{total}} = P_{\text{CHCl}_3} + P_{\text{CH}_2\text{Cl}_2} = 62.2 + 28.58 = 90.78 \, \text{mm Hg}

There might be a minor rounding difference due to arithmetic, hence the closest option is 90.38 mm Hg.

The Correct Option is D

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