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Standard XII

Chemistry

Concentration Terms

In a mixture ...

Question

In a mixture of $N_{2}$ and $C O_{2}$ gases, the partial pressure of $C O_{2}$ is $1.25$ atm . The total pressure mixture is $5 a t m$ . The mol fraction of $N_{2}$ in the mixture is?

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Solution

Given $X_{C O_{2}} = 1.25 a t m$
$∵ X_{N_{2}} + X_{C O_{2}} = 1$
$\Rightarrow X_{N_{2}} = 1 - X_{C O_{2}} = 1 - 1.25 = 0.25$
Total pressure $P_{T} = 5 a t m$
Partial pressure= Mole fraction $\times$ Total pressure
$\Rightarrow P_{N_{2}} = X_{N_{2}} \times P_{T}$
$= 0.25 \times 5$
$= 1.25 a t m$

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In a mixture of N2 and CO2 gases, the partial pressure of CO2 is 1.25 atm . The total pressure mixture is 5atm . The mol fraction of N2 in the mixture is?

Given XCO2=1.25atm∵XN2+XCO2=1⇒XN2=1−XCO2=1−1.25=0.25Total pressure PT=5atmPartial pressure= Mole fraction × Total pressure⇒PN2=XN2×PT =0.25×5 =1.25atm

In a mixture of $N_{2}$ and $C O_{2}$ gases, the partial pressure of $C O_{2}$ is $1.25$ atm . The total pressure mixture is $5 a t m$ . The mol fraction of $N_{2}$ in the mixture is?

Given $X_{C O_{2}} = 1.25 a t m$
$∵ X_{N_{2}} + X_{C O_{2}} = 1$
$\Rightarrow X_{N_{2}} = 1 - X_{C O_{2}} = 1 - 1.25 = 0.25$
Total pressure $P_{T} = 5 a t m$
Partial pressure= Mole fraction $\times$ Total pressure
$\Rightarrow P_{N_{2}} = X_{N_{2}} \times P_{T}$
$= 0.25 \times 5$
$= 1.25 a t m$