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NCERT Class XII Chemistry
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Question : 39 of 53
Marks: +1, -0
The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% and 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K, if the Henry’s law constantsfor oxygen and nitrogen are 3.30×1073.30 \times 10^7 mm Hg and 6.51×1076.51 \times 10^7 mm Hg respectively, calculate the composition of these gases in water.
Solution:  
Total pressure of air in equilibrium with water = 10 atmosphere
As air contains 20% oxygen and 79% nitrogen by volume,
Partial pressure of oxygen (pO2)=20100×10(p_{\mathrm{O}_2}) = \frac{20}{100} \times 10 atm
= 2 atm = 2 × 760 mm Hg = 1520 mm Hg
Partial pressure of nitrogen (pN2)=79100×10(p_{\mathrm{N}_2}) = \frac{79}{100} \times 10 atm
= 7.9 atm = 7.9 × 760 mm Hg = 6004 mm Hg
KH(O2)=3.30×107K_{\mathrm{H}}(\mathrm{O}_2) = 3.30 \times 10^7 mm Hg, KH(N2)=6.51×107K_{\mathrm{H}}(\mathrm{N}_2) = 6.51 \times 10^7 mm Hg
Applying Henry’s law ,pO2=KH×xO2p_{\mathrm{O}_2} = K_{\mathrm{H}} \times x_{\mathrm{O}_2}
or, xO2=pO2KHx_{\mathrm{O}_2} = \frac{p_{\mathrm{O}_2}}{K_{\mathrm{H}}} =1520 mm Hg3.30×107 mm Hg=4.61×10−5= \frac{1520 \,\mathrm{mm\,Hg}}{3.30 \times 10^{7} \,\mathrm{mm\,Hg}} = 4.61 \times 10^{-5}
pN2=KH×xN2p_{\mathrm{N}_2} = K_{\mathrm{H}} \times x_{\mathrm{N}_2}
xN2=pN2KHx_{\mathrm{N}_2} = \frac{p_{\mathrm{N}_2}}{K_{\mathrm{H}}} =6004 mm Hg6.51×107 mm Hg=9.22×10−5= \frac{6004 \,\mathrm{mm\,Hg}}{6.51 \times 10^{7} \,\mathrm{mm\,Hg}} = 9.22 \times 10^{-5}
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