Henry's law constant for the solubility of nitrogen gas in water at `298 K` is `1.0 xx 10^(-5) atm`. The mole fraction of nitrogen in air is `0.8`.The number of moles of nitrogen from air dissolved in `10 mol` of water at `298 K` and `5 atm` pressure is

Options

A. `1 xx 10^(-4)`
B. `2 xx 10^(-4)`
C. `1 xx 10^(-5)`
D. `2 xx 10^(-5)`

Answer

C

Explanation

According to Henery.s law .
`p(N_(2)) = x_(B) xx x(N_(2))`
`0.76 atm = 7.6 xx 10^(4) atm xx x (N_(2))`
`therefore x (N_(2)) = 1 xx 10^(-5)`