Question

If N${}_2$ gas is hubbled through water at 293 K, how many millimoles of N${}_2$ gas would dissolve in 1 litre of water? Assume that N${}_2$ exerts a partial pressure of 0.987 bar. Given that Henry's law constant for N${}_2$ at 293 K is 76.48 kbar.

• A. $0.561$ mmol
• B. $0.222$ mmol
• C. $0.716$ mmol
• D. $0.828$ mmol

Answer 1

Answer: (C)

$0.716$ mmol

According to Henry's law, $P=Kx$

x is the mole fraction
K is the henry's law constant
P is the pressure in atm.

Substitute values in the above expression.

$x=\frac{0.987bar}{76480bar}=1.29\times {10}^{-5}=\frac{n}{55.55}$

$n=7.16\times {10}^{-4}=\frac{7.16\times {10}^{-4}mol\times 1000mmol}{1mol}=0.716mmol$

So, the correct option is $C$