1.

The solubility of a gas in water at 300 K under a pressure of 100 atmospheres is 4xx10^(-3)"kg L"^(-1). Therefore, the mass of the gas in kg dissolved in 250 mL of water under a pressure of 250 atmospheres at 300 K is

Answer»

<P>`2.5xx10^(-3)`
`2.0xx10^(-3)`
`1.25xx10^(-3)`
`5.0xx10^(-3)`

Solution :By HENRY's law`""m=K_(H)p`
at constant temperature, `K_(H)` is constant.
`therefore ` At two different pressures,
`(m)/(m')=(p)/(p')=(4xx10^(-3)"KG L"^(-1))/(m')=(100)/(250)`
`"orm'"=(250)/(100)xx4xx10^(-3)"kg L"^(-1)`
`=10^(-2)"kg L"^(-1)=(10^(-2))/(4)"kg in 250 mL"`
`=2.5xx10^(-3)kg.`


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