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Standard XII

Chemistry

Concentration Terms

If we take 44...

Question

If we take 44g of ${C O}_{2}$ and 14g of $N_{2}$ , what will be the mole fraction of ${C O}_{2}$ in the mixture?

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Solution

The correct option is A $2 / 3$
$X_{C O_{2}} = \frac{n_{C O_{2}}}{n_{C O_{2}} + n_{H_{2}}}$

Mass of $C O_{2}$ = 44 g
Mass pf $N_{2}$ = 14 g

$X_{C O_{2}} = \frac{\frac{44}{44}}{\frac{44}{44} + \frac{14}{28}}$

$\Rightarrow \frac{1}{1 + \frac{1}{2}}$

$\Rightarrow \frac{2}{3}$

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If we take 44g of CO2 and 14g of N2 , what will be the mole fraction of CO2 in the mixture?

The correct option is A 2/3XCO2=nCO2nCO2+nH2Mass of CO2 = 44 gMass pf N2 = 14 gXCO2=44444444+1428⇒11+12⇒23

If we take 44g of ${C O}_{2}$ and 14g of $N_{2}$ , what will be the mole fraction of ${C O}_{2}$ in the mixture?

The correct option is A $2 / 3$
$X_{C O_{2}} = \frac{n_{C O_{2}}}{n_{C O_{2}} + n_{H_{2}}}$

Mass of $C O_{2}$ = 44 g
Mass pf $N_{2}$ = 14 g

$X_{C O_{2}} = \frac{\frac{44}{44}}{\frac{44}{44} + \frac{14}{28}}$

$\Rightarrow \frac{1}{1 + \frac{1}{2}}$

$\Rightarrow \frac{2}{3}$