JEE MAIN - Chemistry (2022 - 24th June Evening Shift - No. 15)
A company dissolves 'x' amount of CO2 at 298 K in 1 litre of water to prepare soda water. X = __________ $$\times$$ 10$$-$$3 g. (nearest integer)
(Given : partial pressure of CO2 at 298 K = 0.835 bar.
Henry's law constant for CO2 at 298 K = 1.67 kbar.
Atomic mass of H, C and O is 1, 12, and 6 g mol$$-$$1, respectively)
Answer
1221
Explanation
According to Henry's law, partial pressure of a gas is given by
$P_{g}=\left(K_{H}\right) X_{g}$
where $X_{g}$ is mole fraction of gas in solution
$0.835=1.67 \times 10^{3}\left(\mathrm{X}_{\mathrm{CO}_{2}}\right)$
$\mathrm{X}_{\mathrm{CO}_{2}}=5 \times 10^{-4}$
Mass of $\mathrm{CO}_{2}$ in $1 \mathrm{~L}$ water $=1221 \times 10^{-3} \mathrm{~g}$
$P_{g}=\left(K_{H}\right) X_{g}$
where $X_{g}$ is mole fraction of gas in solution
$0.835=1.67 \times 10^{3}\left(\mathrm{X}_{\mathrm{CO}_{2}}\right)$
$\mathrm{X}_{\mathrm{CO}_{2}}=5 \times 10^{-4}$
Mass of $\mathrm{CO}_{2}$ in $1 \mathrm{~L}$ water $=1221 \times 10^{-3} \mathrm{~g}$
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