Question

The solubility of a gas in water at $300\, K$ under a pressure of $100$ atmospheres is $4 \times 10^{-3} \, kg / L$. Therefore, the mass of the gas in kilogram dissolved in $250\, mL$ of water under a pressure of $250 \, atm$ and $300\, K$ is:

• A. $2.0 \times 10^{-3}$
• B. $2.5 \times 10^{-3}$
• C. $1.25 \times 10^{-3}$
• D. $5.0 \times 10^{-3}$

Answer 1

Answer: (B)

$\frac{S_{1}}{S_{2}}=\frac{P_{1}}{P_{2}}$
$\frac{4 \times 10^{-3}}{S_{2}}=\frac{100}{250}$
$\therefore S_{2}=\frac{4 \times 10^{-3} \times 250}{100}=10^{-2} \,kg \,L ^{-1}$
Amount of gas dissolved in $250$ water
$=\frac{10^{-2}}{1000} \times 250=2.5 \times 10^{-3} \,kg$