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Standard XII

Chemistry

Empirical & Molecular Formula

The vapour de...

Question

The vapour density of a mixture of gas $A$ (molecular mass = $40$ ) and gas $B$ (molecular mass = $80$ ) is $25$ . Then, mole percentage of gas $B$ in the mixture would be :

25 %

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30 %

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43 %

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45 %

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Solution

The correct option is D $25 %$
As vapour density of mixture is $25$ , Molar mass of mixture is $2 \times 25 = 50$ .
Now, assume mole $%$ of $B$ in mixture is $x$ .
Average mass $= \frac{((100 - x) \times 40 + x \times 80)}{100} = 50$
$∴ x = 25 %$

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The vapour density of a mixture of gas A (molecular mass = 40) and gas B (molecular mass = 80) is 25. Then, mole percentage of gas B in the mixture would be :

The correct option is D 25%As vapour density of mixture is 25, Molar mass of mixture is 2×25=50.Now, assume mole % of B in mixture is x.Average mass =((100−x)×40+x×80)100=50∴x=25%

The vapour density of a mixture of gas $A$ (molecular mass = $40$ ) and gas $B$ (molecular mass = $80$ ) is $25$ . Then, mole percentage of gas $B$ in the mixture would be :

The correct option is D $25 %$
As vapour density of mixture is $25$ , Molar mass of mixture is $2 \times 25 = 50$ .
Now, assume mole $%$ of $B$ in mixture is $x$ .
Average mass $= \frac{((100 - x) \times 40 + x \times 80)}{100} = 50$
$∴ x = 25 %$