Question

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

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

Answer 1

Answer: (A)

As $4 \times 10^{-3} kg$ of gas is dissolved in $1\, L$ of water,
According to Henry' s law,
$m = K _{ H } \rho$
Where, $m =4 \times 10^{-3} kg$
and $\rho=100\, atm$
$\therefore K _{ H }=\frac{ m }{\rho}=\frac{4 \times 10^{-3}}{100}$
$=4 \times 10^{-5} kg L ^{-1}$ atm $^{-1}$
Now, if the pressure is increased to 250 atm the mass of gas
$(m)= K _{ H } P$
$=4 \times 10^{-5} \times 250$
$=10^{-2} kg\, L ^{-1}$
and the mass of gas present in $250\, ml$
$=\frac{250}{100} \times 10^{-2}$
$=2.5 \times 10^{-3} kg$