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CBSE Class 12 Chemistry 2022 Term 1 Paper

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Question : 26 of 55
Marks: +1, -0
SECTION - B
An unknown gas ' XX ' is dissolved in water at 2.5 bar pressure and has mole fraction 0.04 in solution. The mole fraction of ' XX ' gas when the pressure of gas is doubled at the same temperature i
Solution:  
Mole fraction of gas XX in solution =0.04=0.04
Pressure =2.5=2.5 bar
Let
p1  =P0X1p_1\;= P_0 X_1
2.5  =0.04P02.5\;=0.04 P_0 ......(i)
Let pressure be doubled, then p2p_2
5.0=X2P05.0= X_2 P_0 ......(ii)
Dividing Eqn ii by eqn II, we get
5.02.5  =X20.04\frac{5.0}{2.5}\;= \frac{X_2}{0.04}
2×0.04  =X22 \times 0.04\;= X_2
X2  =0.08X_2\;=0.08
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