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Standard XII

Chemistry

Ideal Gas Equation

45 g of gaseo...

Question

$45 g$ of gaseous mixture of $N O_{2}$ & $N_{2} O$ gas occupy $22.7 l i t e r$ at $S T P$ . Find the mole $%$ of $N O_{2}$ in the gaseous mixture.

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Solution

$22.7 L$ is occupied by 1 mole of a substance.
Hence, $45 g$ is the mass of 1 mole or MW of the mixture.
$M W_{m i x} = (X_{N O_{2}}) (M o l e c u l a r w e i g h t o f N O_{2}) + (X_{N_{2} O}) (M o l e c u l a r w e i g h t o f N_{2} O)$
$∵ X_{N O_{2}} + X_{N_{2} O} = 1$
$∴ 45 = (X_{N O_{2}}) (46) + (1 - X_{N_{2} O}) (44)$
$45 - 44 = (46 - 44) X_{N O_{2}}$
$∴ X_{N O_{2}} = \frac{1}{2} = 0.5$
$∴ M o l e %$ of $N O_{2} = % X_{N O_{2}} = 50 %$

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45 g of gaseous mixture of NO2 & N2O gas occupy 22.7 liter at STP. Find the mole % of NO2 in the gaseous mixture.

22.7 L is occupied by 1 mole of a substance.Hence, 45g is the mass of 1 mole or MW of the mixture.MWmix=(XNO2)(Molecularweightof NO2)+(XN2O)(MolecularweightofN2O) ∵XNO2+XN2O=1∴45=(XNO2)(46)+(1−XN2O)(44) 45−44=(46−44)XNO2∴XNO2=12=0.5∴Mole% of NO2=%XNO2=50%

$45 g$ of gaseous mixture of $N O_{2}$ & $N_{2} O$ gas occupy $22.7 l i t e r$ at $S T P$ . Find the mole $%$ of $N O_{2}$ in the gaseous mixture.