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Henry's Law

At 0∘C and 2....

Question

At $0^{\circ} C$ and $2.00 a t m$ partial pressure of oxygen, the aqueous solubility of $O_{2} (g)$ is $60.2 m L$ $O_{2}$ per litre. Thus, molarity of $O_{2}$ in a saturated water solution when the $O_{2}$ is under the normal partial pressure of $0.3024 a t m$ is

4.57 \times 10^{- 4} M

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2.18 \times 10^{- 3} M

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4.57 \times 10^{- 2} M

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8.12 \times 10^{- 4} M

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Solution

The correct option is D $8.12 \times 10^{- 4} M$
Number of mol of $O_{2}$ at $0^{\circ} C$

$= \frac{p V}{R T} = \frac{2.0 a t m \times (\frac{60.2}{1000}) L}{0.0821 L a t m m o l^{- 1} K^{- 1} \times 273 K}$

$= \frac{2 \times 0.0602}{0.0821 \times 273} = \frac{0.1204}{22.4133}$

$= 5.37 \times 10^{- 3} m o l in 1 L H_{2} O$

Thus, molarity of $O_{2}$ gas at $2 a t m$ partial pressure of $O_{2} (g)$ = $5.37 \times 10^{- 3} M$

By Henry's law,

$p$ = $K_{H} x$
Here we have $x$ = molarity of $O_{2} (g)$

$p_{1} = K_{H} x_{1}$

$p_{2} = K_{H} x_{2}$

$∴ \frac{p_{2}}{p_{1}} = \frac{x_{2}}{x_{1}}$

$∴ x_{2} = x_{1} \frac{p_{2}}{p_{1}} = (5.37 \times 10^{- 3} \times \frac{0.3024}{2})$

$x_{2} = 8.12 \times 10^{- 4} M$
Hence, (d) is correct.

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Henry's Law

Standard XII Chemistry

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At 0∘C and 2.00 atm partial pressure of oxygen, the aqueous solubility of O2(g) is 60.2 mL O2 per litre. Thus, molarity of O2 in a saturated water solution when the O2 is under the normal partial pressure of 0.3024 atm is

At $0^{\circ} C$ and $2.00 a t m$ partial pressure of oxygen, the aqueous solubility of $O_{2} (g)$ is $60.2 m L$ $O_{2}$ per litre. Thus, molarity of $O_{2}$ in a saturated water solution when the $O_{2}$ is under the normal partial pressure of $0.3024 a t m$ is