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AIIMS 2019 Chemistry Raoult's Law MCQ Question

Type: MCQ-numerical-Medium-Class 12

The vapour pressure of pure CHCl3\text{CHCl}_3 and CH2Cl2\text{CH}_2\text{Cl}_2 are 200 and 41.5 atm respectively. The weight of CHCl3\text{CHCl}_3 and CH2Cl2\text{CH}_2\text{Cl}_2 are resepectively 11.9 g and 17 gm. The vapour pressure of solution will be

A

80.5

B

79.5

C

94.3

D

105.5

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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