A Mixture of nitrogen and oxygen has a density of $1.00 g / 1$ and the average molecular weight ${(M}_{a y})$ of $31.2 g$ at $27^{o} C$ and $600$ mm Hg per mole pressure. What is the mole fraction of nitrogen?(Assume ideal gas behavior)

0.8

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0.6

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0.4

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0.2

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Solution

The correct option is D $0.2$
$M_{a v g} = X_{N_{2}} \times M_{N_{2}} + X_{O_{2}} \times M_{O_{2}} ∴ 31.2 = X_{N_{2}} \times 28 + X_{O_{2}} \times 32 ⟶ (i)$

Now, PV=nRT
$∴$ $V = \frac{n R T}{P} = \frac{8.314 \times 300}{80000 P a} = 0.0312 m^{3} = 31.2 L i t r e s$

$∴$ Density= $\frac{m a s s}{v o l u m e}$
$∴$ Mass of mixture= $1 \times 31.2 = 31.2 g m$
$∴$ moles of mixture= $\frac{31.2}{31.2} = 1$ mole
$∴$ $X_{O_{2}} + X_{N_{2}} = 1 ⟶ (i i)$

Solving (i) & (ii), we get,
$X_{N_{2}} = 0.2$ & $X_{O_{2}} = 0.8$

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A Mixture of nitrogen and oxygen has a density of 1.00g/1 and the average molecular weight (May) of 31.2g at 27oC and 600mm Hg per mole pressure. What is the mole fraction of nitrogen?(Assume ideal gas behavior)

The correct option is D 0.2Mavg=XN2×MN2+XO2×MO2∴31.2=XN2×28+XO2×32⟶(i)Now, PV=nRT∴ V=nRTP=8.314×30080000Pa=0.0312m3=31.2Litres∴ Density= massvolume∴ Mass of mixture= 1×31.2=31.2gm∴ moles of mixture= 31.231.2=1 mole∴ XO2+XN2=1⟶(ii)Solving (i) & (ii), we get,XN2=0.2 & XO2=0.8

A Mixture of nitrogen and oxygen has a density of $1.00 g / 1$ and the average molecular weight ${(M}_{a y})$ of $31.2 g$ at $27^{o} C$ and $600$ mm Hg per mole pressure. What is the mole fraction of nitrogen?(Assume ideal gas behavior)