1.

The Henry's law constant for the solubility of N_(2) gas in water at 298 K is 1.0 xx 10^(5) atm. The mole fraction of N_(2) in air is 0.8. The number of moles of N_(2) from air dissolved in 10 moles of water at 298 K and 5 atm. Pressure is:

Answer»

`4.0xx10^(-4)`
`4.0xx10^(-5)`
`5.0xx10^(-4)`
`4.0xx10^(-6)`

Solution :`P_(N_(2))=K_(H)x_(N_(2)) ("in air")`
But `P_(N_(2))=X_(N_(2))xxPT=0.8xx5` ATM (in WATER)
`(4 atm)= K_(H)x_(N_(2))`
`x_(N_(2))=((4atm))/K_(H)=((4atm))/((1.0xx10^(5)atm))`
`=4xx10^(-5)`
`"Now" n_(N_(2))/(n_(H_(2))+n_(H_(2)O))=x_(N_(2))=4xx10^(-5)`
`n_(H_(2))/n_(H_(2)O)=4xx10^(-5)`
`n_(H_(2))=4xx10^(-5)xxn_(H_(2)O)`
` 4xx10^(-5)XX(10 "mol")`
`=4xx10^(-4)"mol"`


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