Question

Calculate the amount of oxygen at $0.20$ atm dissolved in $1$ kg of water at $293$ K. The Henry's law constant for oxygen is $4.58$ $\times$ 10${}^4$ atmosphere at $293$ K.

• A. $7.07$ $\times$ $10$${}^{-6}$ kg
• B. $7.27$ $\times$ ${10}^{-6}$ kg
• C. $7.57$ $\times$ ${10}^{-6}$ kg
• D. $7.77$ $\times$ ${10}^{-6}$ kg

Answer 1

The correct option is D $7.77$ $\times$ ${10}^{-6}$ kg

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:

$x=\frac{0.20atm}{4.58\times {10}^{4}atm}=0.04\times {10}^{-4}=\frac{n}{55.55}$

$n=2.43\times {10}^{-4}$

Thus $2.43\times {10}^{-4}$ moles of oxygen are dissolved in 1 L of water or 1 kg of water.

This corresponds to $2.43\times {10}^{-4}\times 32=77.7\times {10}^{-4}g=7.762\times {10}^{-6}kg$ of oxygen.