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N(2) gas is bubbled through water at 293...

`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

A

0.129 nV/mol

B

0.716 m/mol

C

1.29 m/mol

D

7.16 m/mol

Text Solution

Verified by Experts

The correct Answer is:
B
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
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Knowledge Check

  • 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 100 kbar. If the gas exerts a partial pressure of 1 bar, the number of millimoles of the gas dissolved in one litre of water is

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