Question

The Henry's law constant for the solubility of $N_2$ gas in water at 298 K is $1.0\times {10}^{5}atm$. The mole fraction of $N_2$ in air is 0.8. The no. of moles of $N_2$ in 10 moles of water of 298 K and 5 atm pressure is:

• A. $2\times {10}^{-5}$
• B. $6\times {10}^{-5}$
• C. $4\times {10}^{-4}$
• D. $9\times {10}^{-6}$

Answer 1

The correct option is C $4\times {10}^{-4}$

Total pressure is 5 atm. The mole fraction of nitrogen is 0.8.

$Hence,thepartialpressureofnitrogen=P_T\times X=0.8\times 5=4atm$

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.

$X=\frac{4atm}{1\times {10}^{5}atm}=4\times {10}^{-5}$
The no. of moles of $N_2$ in 10 moles of water is-

$\frac{n}{10}=4\times {10}^{-5}$

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