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Question:
$\mathrm{CO}_{2}$ gas is bubbled through water during a soft drink manufacturing process at $298 \mathrm{~K}$. If $\mathrm{CO}_{2}$ exerts a partial pressure of $0.835$ bar then $\mathrm{x} \mathrm{m} \mathrm{mol}$ of $\mathrm{CO}_{2}$ would dissolve in $0.9 \mathrm{~L}$ of water. The value of $x$ is - (Nearest integer)

(Henry's law constant for $\mathrm{CO}_{2}$ at $298 \mathrm{~K}$ is $1.67 \times 10^{3}$ bar $)$

Solution:

From Henry's law

$\mathrm{P}_{\mathrm{gas}}=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{X}_{\mathrm{gas}}$

$0.835=1.67 \times 10^{3} \times \frac{\mathrm{n}\left(\mathrm{CO}_{2}\right)}{\frac{0.9 \times 1000}{18}}$

$\mathrm{n}\left(\mathrm{CO}_{2}\right)=0.025$

Millimoles of $\mathrm{CO}_{2}=0.025 \times 1000=25$

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