`N_(2)` gas is bubbled through water at `293 K` and the partial pressure of `N_(2)` is `0.987` bar .If the henry's law constant for `N_(2)` at `293 K` is `76.84` kbar, the number of millimoles of `N_(2)` gas that will dissolve in `1 L` of water at `293 K` is

Options

A. 0.129 nV/mol
B. 0.716 m/mol
C. 1.29 m/mol
D. 7.16 m/mol

Answer

B

Explanation

According to Henry's Law
`P_(N_(2))=K_(H)xx chi_(N_(2))`
`chi_(N_(2))=(p_(N_(2)))/(K_(H))=(0.987 "bar")/(76480 "bar")=1.29xx10^(-5)`
`n` moles of `N_(2)` are present in `1L` of water `chi_(N_(2))=n/(n+55.5)=1.29xx10^(-5)`
`:.n lt lt lt 55.5`
`:.n=71.595xx10^(-5)` mol
`=0.716` m/mol