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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 $25^{o}$ & 1 atm will be if air is 20% by volume of $O_{2}$ and 80% by volume of $N_{2}$ .
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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1 : 2

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2 : 1

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Solution

The correct option is D $2 : 1$
$X_{N_{2}} = 0.8, X_{O_{2}} = 0.2$ 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. At

20^{o} C

and a pressure of 1 atm each gas, 1 litre of water dissolves 0.043 g of pure

O_{2}

or 0.019 g of pure

N_{2}

. Assuming that dry air is composed of 20%

O_{2}

and 80%

N_{2}

(by volume) determine the masses of

O_{2}

and

N_{2}

dissolved by 1 litre of water at

20^{o} C

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

H_{2}

20^{o} C

be 'b' atm

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

298 K

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

What will be the moles of

N_{2}

1 k g H_{2} O ?

Δ G = 77.77 k J {m o l}^{- 1}

1000 K

for the reaction

\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) = N O (g)

. Partial pressure of

N O

under equilibrium at

1000 K

for air at 1 atm pressure containing 80%

N_{2}

and 20%

O_{2}

by volume is

X \times 10^{- 5}

atm.

100 X

is__________.

Q. Air contains

N_{2}

and

O_{2}

4 : 1

ratio by volume. Approximate molecular weight of air (g/mole) is:

Q. Only

N_{2}

, CO and

C O_{2}

gases remain after 0.72 gm of carbon is treated with one litre of air at

27^{0} C

and 4.92 atm pressure. Assume air composition

O_{2}

= 20%,

N_{2}

= 79% and

C O_{2}

= 1% (by volume). The heat evolved (in Kcal) under constant pressure is:
Given :

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

C + \frac{1}{2} O_{2} \to C O, Δ H = - 25 K c a l / m o l

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

The correct option is D 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 $25^{o}$ & 1 atm will be if air is 20% by volume of $O_{2}$ and 80% by volume of $N_{2}$ .
Given : $K_{H} (N_{2}) = 2 \times 10^{4} a t m$
$K_{H} (O_{2}) = 10^{4} a t m$

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