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NVAMediumJEE 2023Raoult's Law & Vapour Pressure

JEE Chemistry 2023 Question with Solution

Let V.P. of pure BB be P0P_0. When XA=0.7X_A = 0.7 and XB=0.3X_B = 0.3, the Total pressure is 350 mm Hg350 \, \text{mm Hg}. If XA=0.2X_A = 0.2 and XB=0.8X_B = 0.8, the total pressure becomes 410 mm Hg410 \, \text{mm Hg}. Calculate the vapor pressure of AA and BB.

Answer

Correct answer:314

Step-by-step solution

Standard Method

Given: For an ideal solution,

Ptotal=PA0XA+PB0XBP_{\text{total}} = P_A^0 X_A + P_B^0 X_B

First mixture: Ptotal=350 mm HgP_{\text{total}} = 350 \, \text{mm Hg}, XA=0.7X_A = 0.7, XB=0.3X_B = 0.3.

Second mixture: Ptotal=410 mm HgP_{\text{total}} = 410 \, \text{mm Hg}, XA=0.2X_A = 0.2, XB=0.8X_B = 0.8.

Find: Vapour pressures of pure AA and pure BB.

Using Raoult's law for the first mixture,

350=PA0⋅0.7+PB0⋅0.3350 = P_A^0 \cdot 0.7 + P_B^0 \cdot 0.3

For the second mixture,

410=PA0⋅0.2+PB0⋅0.8410 = P_A^0 \cdot 0.2 + P_B^0 \cdot 0.8

Elimination and Substitution

From the first equation,

PB0=350−PA0⋅0.70.3P_B^0 = \frac{350 - P_A^0 \cdot 0.7}{0.3}

Substitute this into the second equation:

410=PA0⋅0.2+(350−PA0⋅0.70.3)⋅0.8410 = P_A^0 \cdot 0.2 + \left( \frac{350 - P_A^0 \cdot 0.7}{0.3} \right) \cdot 0.8

Simplifying,

410=PA0⋅0.2+280−PA0⋅0.560.3410 = P_A^0 \cdot 0.2 + \frac{280 - P_A^0 \cdot 0.56}{0.3} 410=PA0⋅0.2+2800.3−PA0⋅0.560.3410 = P_A^0 \cdot 0.2 + \frac{280}{0.3} - \frac{P_A^0 \cdot 0.56}{0.3} 410=PA0⋅0.2+933.33−1.87PA0410 = P_A^0 \cdot 0.2 + 933.33 - 1.87 P_A^0 410=933.33−1.67PA0410 = 933.33 - 1.67 P_A^0 PA0=933.33−4101.67P_A^0 = \frac{933.33 - 410}{1.67} PA0=314 mm HgP_A^0 = 314 \, \text{mm Hg}

Therefore, the vapour pressure of pure AA is 314 mm Hg314 \, \text{mm Hg}. The solution concludes with the numerical answer 314314.

Common mistakes

  • Using mole fractions incorrectly. XAX_A and XBX_B must be paired with PA0P_A^0 and PB0P_B^0 respectively in Raoult's law. Swapping them gives wrong simultaneous equations.

  • Assuming the asked numerical answer must include both vapour pressures. The provided solution extracts only PA0=314 mm HgP_A^0 = 314 \, \text{mm Hg} as the final numerical result, so the accepted NVA answer is 314314.

  • Treating total pressure as a simple average of pure vapour pressures. For an ideal solution, total pressure is the mole-fraction-weighted sum, not an arithmetic mean.

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