1.

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

Answer»

`1:4`
`4 : 1`
`1 : 2`
`2 : 1`

Solution :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 `


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