1.

The mole fraction of a gas dissolved in a solvent is given by Henry's law. Constant for gas in water at 298 K is 5.55xx10^(7) Torr and the partial pressure of the gas is 200 Torr, then what is the amount of the gas dissolved in 1.0 kg of water?

Answer»

`2.0xx10^(-4)mol`
`2.5xx10^(-5)mol`
`3.7xx10^(-6)mol`
`1.2xx10^(-8)mol`

SOLUTION :By Henry's law, `p_(A)=K_(H)XX x_(A)`
`"or"x_(A)=(p_(A))/(K_(H))=("200Torr")/(5.55xx10^(7)"Torr")=3.6xx10^(-6)`
`"But"x_(A)=(n_(A))/(n_(A)+n_(H_(2)O))~=(n_(A))/(n_(H_(2)O))=(n_(A))/(1000//18)`
`therefore""n_(A)=m_(A)xx(1000)/(18)=3.6xx10^(-6)xx(1000)/(18)" mole"`
`=2.0xx10^(-4)"mole."`


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