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 x-10-5 mol dm-3- The value of xRound off to the Nearest Integer is -Given -Henry-s law constantKH -8-0- 104kPa for O2Density of water with dissolved oxygen-1-0 Kg dm-3 -

P_O_2(over water) = 20 nbsp; kPa K_H for O_2=8.0× 10^4kPa If nbsp; χ_o_2 is the mole fraction of nbsp; O_2 in solution , then according to Henry's law ...

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Standard XII

Chemistry

Gases in Liquids

The oxygen di...

Question

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 $x \times 10^{- 5} m o l d m^{- 3}$ . The value of x
(Round off to the Nearest Integer ) is
$[G i v e n : Henry's law constant (K_{H}) = 8.0 \times 10^{4} k P a for O_{2} Density of water with dissolved oxygen = 1.0 K g d m^{- 3}]$

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Solution

$P_{O_{2}}$ (over water) = 20 kPa

$K_{H} f o r O_{2} = 8.0 \times 10^{4} k P a$

If $χ_{o_{2}}$ is the mole fraction of $O_{2}$ in solution , then according to Henry's law

$P_{o_{2}} = K_{H} (χ_{O_{2}})$

$χ_{O_{2}} = \frac{20}{8.0 \times 10^{4}} = 2.5 \times 10^{- 4}$
Since density of water is 1 kg/L
$Mass of 1 Kg of water containing O_{2} = 1 L$

$∴ Molarity of O_{2} in solution = 25 \times 10^{- 5} M x = 25$

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PO2(over water) = 20 kPa KH for O2=8.0×104kPa If χo2 is the mole fraction of O2 in solution , then according to Henry's law Po2=KH(χO2) χO2=208.0×104=2.5×10−4 Since density of water is 1 kg/L Mass of 1 Kg of water containing O2=1 L ∴Molarity of O2 in solution =25×10−5Mx=25