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 `0.6`. The number of moles of `N_(2)` from air dissolved in 10 moles of water at `298 K`and `5 atm` pressure is
a. `3.0 xx 10^(-4)` b. `4.0 xx 10^(-5)` c. `5.0 xx 10^(-4)` d.`6.0 xx 10^(-6)`

Options

A. `3.0xx10^(-4)`
B. `4.0xx10^(-5)`
C. `5.0xx10^(-4)`
D. `6.0xx10^(-6)`

Answer

A

Explanation

According to Henry.s law, `P_(N_(2))=K_(H), X_(N_(2))`
i.e., `x_(N_(2))=(P_(N_(2)))/(K_(H))`