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At 20^(@)C, the solubility of N(2) in wa...

At `20^(@)C`, the solubility of `N_(2)` in waer is 0.0150 `gL^(-1)` when the partial pressure of `N_(2)` is 580 mm Hg. What will be the solubility of `N_(2)` in water at `20^(@)C` when its partial pressure is 800 mm Hg ?

Text Solution

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The correct Answer is:
0.0207 gL^(-1)`
`C_(1)=0.0150 gL^(-1), C_(2)=?`
`P_(1)=580mm Hg, P_(2)=800 mmHg`
According to Henry'a Law,
`C_(2)/C_(1)=P_(2)/P_(1)or C_(2)=P_(2)/P_(1)xxC_(1)`
`C_(2)=((800mm))/((580mm))xx(0.0150 gL^(-1))=0.0207 gL^(-1).`
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