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AIIMS 2019 Chemistry Raoult's Law MCQ Question

Type: MCQ-numerical-Medium-Class 12

The vapour pressure of pure CHCl₃ and CH₂Cl₂ are 200 and 41.5 atm respectively. The weight of CHCl₃ and CH₂Cl₂ are respectively 11.9 g and 17 gm. The molar mass of CHCl₃ is 119.5 g/mol and the molar mass of CH₂Cl₂ is 85 g/mol. The number of moles of CHCl₃ is calculated below.

Correct Answer

Option C

Detailed Explanation

To calculate the number of moles of CHCl₃, we use the formula:

Number of moles=mass (g)molar mass (g/mol)\text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}

Substituting the values for CHCl₃, we have:

Number of moles of CHCl₃=11.9 g119.5 g/mol≈0.0995 mol\text{Number of moles of CHCl₃} = \frac{11.9 \, \text{g}}{119.5 \, \text{g/mol}} \approx 0.0995 \, \text{mol}

This calculation confirms that option C is correct. Other options are not applicable as they do not provide relevant information or calculations related to the number of moles of CHCl₃. Understanding this calculation is essential for solving problems involving vapor pressure and colligative properties in solutions.

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