When a gas is bubbled through water at 298 K, a very dilute solution of the gas is obtained. Henry's law constant for the gas at 298 K is 100kbar. If the gas exerts a partial pressure of 1 bar, the number of millimoles of the gas dissolved in one litre of water is

Options

A. `0.555`
B. `5.55`
C. `0.0555`
D. `55.5`

Answer

A

Explanation

`k_(H) = 100` kbar, p = 1 bar
`p=k_(H)xx` mole fraction
Mole fraction `=(p)/(k_(H))=(1)/(100xx10^(3))=10^(-5)`
Mole fraction = `("Moles of gas")/("Total moles")`
Moles of water = `("Weight of water")/("Molecular weight of water")`
Weight of water = 1000 g (`because` 1000 mL = 1000 g)
Moles `=(1000)/(18)=55.5`
Mole fraction = `10^(-5)=(x)/(55.5+x)`
As `55.5gtgtgtx`, thus neglecting x from denominator
`10^(-5)=(x)/(55.5)impliesx=55.5xx10^(-5)` moles
or `55.5xx10^(-2)` millimoles = 0.555 millimoles.