The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is \_\_\_\_ \(\times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3}\). (Round off to the nearest integer). [ Given : Henry's law constant \(=\mathrm{K}_{\mathrm{H}}=8.0 \times 10^{4} \mathrm{kPa}\) for \(\mathrm{O}_{2}\). \[ \text { Density of water with dissolved oxygen } \left.=1.0 \mathrm{~kg} \mathrm{dm}^{-3}\right] \]

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Solutions

The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water.
The molar solubility of oxygen in water is ____

\times 1 0^{- 5} mol dm^{- 3}

. (Round off to the nearest integer).
[ Given : Henry's law constant

= K_{H} = 8.0 \times 1 0^{4} kPa

for

O_{2}

Density of water with dissolved oxygen = 1.0 kg dm^{- 3}]

Answer: 1389

Henry's law fixes the mole fraction of dissolved oxygen:

x = \frac{p}{K _{H}} = \frac{20}{8.0 \times 1 0 ^{4}} = 2.5 \times 1 0^{- 4}

One cubic decimetre of the solution weighs

1.0

kg and is essentially water, so it contains

\frac{1000}{18} = 55.56

mol of water. The solution is very dilute, so the mole fraction may be taken against the water alone and the amount of oxygen in that volume is

n = x \times 55.56 = 2.5 \times 1 0^{- 4} \times 55.56 = 1.389 \times 1 0^{- 2} mol

The molar solubility of oxygen is therefore

1389 \times 1 0^{- 5}

mol dm

^{- 3}

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Integer answer