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CBSE Class 12 Chemistry 2020 Outside Delhi Set 1 Solved Paper

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Question : 27 of 37
Marks: +1, -0
State Henry's law. Calculate the solubility of CO2\mathrm{CO}_2 in water at 298 K298\,\mathrm{K} under 760 mm Hg760\,\mathrm{mm}\,\mathrm{Hg}.
( KH\mathrm{K}_{\mathrm{H}} for CO2\mathrm{CO}_2 in water at 298 K298\,\mathrm{K} is 1.25×106 mm Hg1.25 \times 10^6\,\mathrm{mm}\,\mathrm{Hg} )
Solution:  
Henry's law: The mass of a gas dissolved in a given volume of the liquid at a constant temperature depends upon the pressure which is applied.
KH   for   CO2=1.25×106 mm Hg  \mathrm{K}_{\mathrm{H}}\;\text{ for }\;\mathrm{CO}_2 = 1.25 \times 10^6\,\mathrm{mm}\,\mathrm{Hg}\;
xCO2=  Partial pressure of  CO2KH  for  CO2x_{\mathrm{CO}_2} = \frac{\;\text{Partial pressure of}\;\mathrm{CO}_2}{\mathrm{K}_{\mathrm{H}}\;\text{for}\;\mathrm{CO}_2} =760 mm Hg1.25×106 mm Hg.= \frac{760\,\mathrm{mm}\,\mathrm{Hg}}{1.25 \times 10^6\,\mathrm{mm}\,\mathrm{Hg}.}
  =608×10−6\;=608 \times 10^{-6}
Mole fraction represents the solubility of CO2\mathrm{CO}_2 in water.
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