Question

The Henry's law constant for the solubility of $N_2$ gas in water at $298K$ is $1.0\times 105atm$. The mole fraction of $N_2$ in air is $0.8$. The number of moles of $N_2$ from air dissolved in $10$ moles of water at $298K$ and $5atm$ pressure is

• A. $4\times {10}^{-4}$
• B. $4\times {10}^{-5}$
• C. $5\times {10}^{-4}$
• D. $4\times {10}^{-6}$

Answer 1

Answer: (A)

$4\times {10}^{-4}$

The total pressure is 5 atm. The mole fraction of Nitrogen is 0.8.

Hence, the partial pressure of Nitrogen = mole fraction of Nitrogen × Total pressure = $0.8\times 5=4atm$

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.

$x=\frac{4atm}{1\times {10}^{5}atm}=4\times {10}^{-5}=\frac{n}{n+10}$

where n is no. of mole of nitrogen

$n=4.0\times {10}^{-4}mol$

So, the correct option is $A$