Question

1 kg of water under a nitrogen pressure of 1 atmosphere dissolves $0.02$ gm of nitrogen at $293$ K.

Calculate Henry's law constant.

• A. $7.2\times {10}^{-4}$ atm
• B. $7.5\times {10}^4$ atm
• C. $7.4\times {10}^4$ atm
• D. $7.2\times {10}^4$ atm

Answer 1

The correct option is A $7.2\times {10}^{-4}$ atm

For water, $1kg=1L$
0.02 g nitrogen corresponds to $\frac{0.02}{28}=7.2\times {10}^{-4}$moles.
Hence, the solubility $S=7.2\times {10}^{-4}M$
According to Henry's law, $S=KP$
S is the solubility in moles per litre
K is the henry's law constant
P is the pressure in atm.
Substitute values in the above expression.
$7.2\times {10}^{-4}=K\times 1atm$
$K=7.2\times {10}^{-4}Lat{m}^{-1}$