1.

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 ?

Answer»


Solution :`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).`


Discussion

No Comment Found

Related InterviewSolutions