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Standard XII

Chemistry

Gases in Liquids

Ratio of solu...

Question

Ratio of solubilities of gases $N_{2}$ & $O_{2}$ in water from air at $25^{o}$ and 1 atm will be if air is 20% by volume of $O_{2}$ and 80% by volume of $N_{2}$ is:
Given : $K_{H} (N_{2}) = 2 \times 10^{4} a t m$
$K_{H} (O_{2}) = 10^{4} a t m$

8 : 1

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1 : 8

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2 : 1

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1 : 2

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Solution

The correct option is C 2 : 1
$X_{N_{2}} = 0.8, X_{O_{2}} = 0.2$ in air

$∴ P_{N_{2}} = 0.8 a t m, P_{O_{2}} = 0.2 a t m$

$P = K_{H} X_{4}$ where $X$ is solubility

$0.8 = 2 \times 10^{4} X_{N_{2}}$ .......(1)

$0.2 = 10^{4} X_{O_{2}}$ .......(2)

Ratio of solubility of $N_{2}$ & $O_{2}$ is $2 : 1$

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Similar questions

Q. The major components of air are

O_{2}

and

N_{2}

with approximate proportion of 21% and 79% by volume at

298 K

. The water is equilibrium with air at a pressure of

10 a t m

. At

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K_{H} (N_{2}) = 6.51 \times 10^{7} m m

What will be the moles of

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1 k g H_{2} O ?

Q. At

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or 0.019 g of pure

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and 80%

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(by volume) determine the masses of

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exposed to air at a total pressure of 1 atm. Assume V.P. of

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be 'b' atm

Q. Only

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Given :

C + O_{2} \to C O_{2}, Δ H = - 100 K c a l / m o l

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Q. Air contains

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1000 K

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. Partial pressure of

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Ratio of solubilities of gases N2 & O2 in water from air at 25o and 1 atm will be if air is 20% by volume of O2 and 80% by volume of N2 is:Given : KH(N2)=2×104atm KH(O2)=104atm

The correct option is C 2 : 1XN2=0.8,XO2=0.2 in air∴PN2=0.8atm,PO2=0.2atmP=KHX4 where X is solubility0.8=2×104XN2 .......(1)0.2=104XO2 .......(2)Ratio of solubility of N2 & O2 is 2:1

Ratio of solubilities of gases $N_{2}$ & $O_{2}$ in water from air at $25^{o}$ and 1 atm will be if air is 20% by volume of $O_{2}$ and 80% by volume of $N_{2}$ is:
Given : $K_{H} (N_{2}) = 2 \times 10^{4} a t m$
$K_{H} (O_{2}) = 10^{4} a t m$