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Air contains O_(2) and N_(2) in the ratio of 1:4. Calculate the ratio of solubilities in terms of mole fraction of O_(2) and N_(2) dissolved in water at atmospheric pressure and room temperature at which Henry's constant for O_(2) and N_(2) are 3.30xx10^(7) torr and 6.60xx10^(7) torr respectively. |
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Answer» Solution :At 1 BAR pressure, Partial pressure of `O_(2)(p_(O_(2)))=(1)/(5)xx"1 bar"="0.2 bar,Partial pressure of "N_(2)(p_(N_(2)))=(4)/(5)xx"1 bar= 0.8 bar"` `{:("Applying Henry's law,",p_(O_(2))=K_(H)(O_(2))xxx_(O_(2)),or,x_(O_(2))=(p_(O_(2)))/(K_(H)(O_(2)))),(,p_(N_(2))=K_(H)(N_(2))xx x_(N_(2)),or, x_(N_(2))=(p_(N_(2)))/(K_(H)(N_(2)))):}` `therefore (x_(O_(2)))/(x_(N_(2)))=(p_(O_(2)))/(K_(H)(O_(2)))xx(k_(H)(N_(2)))/(p_(N_(2)))=(p_(O_(2)))/(p_(N_(2)))xx(K_(H)(N_(2)))/(K_(H)(O_(2)))=("0.2 bar")/("0.8 bar")xx(6.60xx10^(7)" TORR")/(3.30 xx10^(7)" torr")=(1)/(2), i.e., x_(O_(2)):x_(N_(2))=1:2` |
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