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JEE Advanced 2009Paper 1
Exam
JEE Advanced
Year
2009
Paper
Paper 1
Subject
Chemistry
Chapter
Solutions

JEE Advanced 2009 · Paper 1 · Chemistry · Solutions

The Henry's law constant for the solubility of N_2 gas in water at...

The Henry's law constant for the solubility of N2\mathrm{N}_2 gas in water at 298 K298 \mathrm{~K} is 1.0×105 atm1.0 \times 105 \mathrm{~atm}. The mole fraction of N2\mathrm{N}_2 in air is 0.80.8. The number of moles of N2\mathrm{N}_2 from air dissolved in 10 moles of water of 298 K298 \mathrm{~K} and 5 atm5 \mathrm{~atm} pressure is

Options

  1. A4×10−44 \times 10^{-4}
  2. B4.0×10−54.0 \times 10^{-5}
  3. C5.0×10−45.0 \times 10^{-4}
  4. D4.0×10−64.0 \times 10^{-6}

Correct answer

A. 4×10−44 \times 10^{-4}

Step-by-step solution

PN2=KH×P_{\mathrm{N}_2}=K_{\mathrm{H}} \times mole-fraction (N2)\left(\mathrm{N}_2\right) mole-fraction N21105×0.8×5=4×10−5mol−1In10molesolubility\mathrm{N}_{2} \frac{1}{10^5} \times0.8 \times5=4 \times10^{-5} \mathrm{ mol}^{-1}In 10 mole solubility in 4×10−44 \times10^{-4}.
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