If N 2 gas is bubbled through water at 293 K, how many millimoles of N 2 gas would dissolve in 300 mole of water, if N 2 exerts a partial pressure of 1 bar. Given that Henry's law constant for N 2 at 293 K is 75.00 kbar.
Text Solution
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4 mmol.
The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry's law. Thus :
x (Nitrogen) = 
= 
if n represents number of moles of N 2 in solution,
x (Nitrogen) = 
= 
= 
(n in denominator is neglected as it is < < 300)
Thus n = 4 × 10 –3 mol. = 4 mmol
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