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Standard XII

Chemistry

Concentration Terms

A gas 'X' i...

Question

A gas $^{'} X^{'}$ is dissolved in water at $^{'} 2^{'}$ bar pressure. Its mole fraction is $0.02$ in solution. The mole fraction of water when the pressure of gas is doubled at the same temperature is:

0.04

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0.98

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0.96

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0.02

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Solution

The correct option is B $0.96$
$P_{1} = P_{o} X_{1}$
$2 = 0.02 P_{o}$
Let the pressure be doubled , $P_{2}$
then, $4 = P_{o} X_{2}$
so we can write that :
$\frac{P_{1}}{P_{2}} = \frac{X_{1}}{X_{2}}$
$\frac{2}{4} = \frac{0.02}{X_{2}}$
$X_{2} = 0.02 \times \frac{4}{2} = 0.04$
now,
$X w a t e r = 1 - X s o l u t e$
$X w a t e r = 1 - X s o l u t e = 0.04 = 0.96$

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A gas ′X′ is dissolved in water at ′2′ bar pressure. Its mole fraction is 0.02 in solution. The mole fraction of water when the pressure of gas is doubled at the same temperature is:

The correct option is B 0.96P1=PoX12=0.02PoLet the pressure be doubled , P2then, 4=PoX2so we can write that : P1P2=X1X224=0.02X2 X2=0.02×42=0.04now, Xwater=1−XsoluteXwater=1−Xsolute=0.04=0.96

A gas $^{'} X^{'}$ is dissolved in water at $^{'} 2^{'}$ bar pressure. Its mole fraction is $0.02$ in solution. The mole fraction of water when the pressure of gas is doubled at the same temperature is: