Henry’s law constant of oxygen is 1.4×10−3mol.lit−1⋅atm−1at 298 K. How much of oxygen is dissolved in 100 ml at 298 K when the partial pressure of oxygen is 0.5 atm?

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Henry’s law constant of oxygen is $1 .4 \times 10^{- 3} mol . {lit}^{- 1} \cdot {atm}^{- 1}$ at 298 K. How much of oxygen is dissolved in 100 ml at 298
K when the partial pressure of oxygen is 0.5 atm?

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1.4 g

3.2 g

22.4 g

2.24 mg

answer is D.

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Detailed Solution

$given K_{H} = 1 .4 \times 10^{- 3} p_{O_{2}} = 0 .5 or p_{O_{2}} = K_{H} \times x_{O_{2}} ∴ x_{O_{2}} = \frac{0 .5}{1 .4 \times 10^{- 3}} No. of moles; n = \frac{m}{M} 0 .7 \times 10^{- 4} = \frac{m}{32} m = 22 .4 \times 10^{- 4} g = 2 .24 mg$