Air contain `O_(2)` and `N_(2)` in the ration of 1 : 4 The Henry constant for `O_(2)` and `N_(2)` are `3.30 xx 10^(7)` torr and `6.60 xx 10^(7)` torr respectively. Ration of solubility of `O_(2)` and `O_(2)` dissolved in water at atomspheric pressure and room temperature is

Options

A. `1:4`
B. `4 : 1`
C. `1 : 2`
D. `2 : 1`

Answer

C

Explanation

At 1 bar, `Po_(2)=1/4 xx1 "bar"=0.2 "bar",`
`P_(N2)=4/5 xx 1 "bar" = 0.8 "bar"`
A/c. to Henry's law.
`Xo_(2)=(Po_(2))/(K_(H)(o_2)) "and"X_(N2)=P_(N2)/(K_(H)(N_(2)))`
`(Xo_(2))/X_(N2)=P_(O2)/(K_(H)(o2)) xx (K_(H)(N_(2)))/P_(N2)=P_(o2)/P_(N2)xx (K_(H)(N2))/(K_(H)(o_(2)))`
`0.2/0.8 xx (6.60 xx 10^(7))/(3.30 xx 10^(7))=1/2`
`X_(O2) : X_(N2) rArr S_(O2) : S_(N2) =1 : 2 `