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MCQEasyJEE 2026Raoult's Law & Vapour Pressure

JEE Chemistry 2026 Question with Solution

At 298 K298 \, \text{K}, the mole percentage of N2(g)N_2\text{(g)} in air is 80%80\%. Water is in equilibrium with air at a pressure of 10 atm10 \, \text{atm}. What is the mole fraction of N2(g)N_2\text{(g)} in water at 298 K298 \, \text{K}? (KHK_H for N2N_2 = 6.5×107 mm Hg6.5 \times 10^7 \, \text{mm Hg})

  • A

    9.35×10−59.35 \times 10^{-5}

  • B

    1.17×10−41.17 \times 10^{-4}

  • C

    9.35×1059.35 \times 10^{5}

  • D

    1.23×10−71.23 \times 10^{-7}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Mole percentage of N2N_2 in air = 80%80\%, total pressure = 10 atm10 \, \text{atm}, and KH=6.5×107 mm HgK_H = 6.5 \times 10^7 \, \text{mm Hg}.

Find: Mole fraction of N2N_2 dissolved in water.

Step 1: Partial pressure of N2N_2

PN2=0.8×10=8 atmP_{N_2} = 0.8 \times 10 = 8 \, \text{atm}

Step 2: Using Henry's law

x=PKHx = \frac{P}{K_H}

Convert pressure into mm Hg:

8×760=6080 mm Hg8 \times 760 = 6080 \, \text{mm Hg}

Now substitute:

x=60806.5×107=9.35×10−5x = \frac{6080}{6.5 \times 10^7} = 9.35 \times 10^{-5}

Therefore, the mole fraction of N2N_2 in water is 9.35×10−59.35 \times 10^{-5}. The correct option is A.

Common mistakes

  • Using the total pressure 10 atm10 \, \text{atm} directly in Henry's law is incorrect because only the partial pressure of N2N_2 should be used. First calculate PN2=0.8×10P_{N_2} = 0.8 \times 10.

  • Not converting pressure from atm to mm Hg is incorrect because KHK_H is given in mm Hg. Keep pressure units consistent before substitution.

  • Interpreting higher Henry's constant as higher solubility is incorrect. A larger KHK_H means lower solubility, so the mole fraction should be very small.

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