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Standard XII

Chemistry

Mole Fraction

A mixture of ...

Question

A mixture of $O_{2}$ and gas "Y" (mol.wt.80) in the mole ratio $a : b$ has a mean molecular weight 40. What would be mean molecular weight, if the gases are mixed in the ratio $b : a$ under identical conditions? (gases are non-reacting):

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Solution

The correct option is D 72
Let mole fraction of $O_{2}$ is $x$
$40 = 32 \times x + 80 (1 - x)$
Or $x = \frac{5}{6}$
$a : b = x : (1 - x) = \frac{5}{6} : \frac{1}{6}$
When ratio is changed
$M_{m i x t u r e} = 32 \times \frac{1}{6} + 80 \times \frac{5}{6} = 72$

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A mixture of O2 and gas "Y" (mol.wt.80) in the mole ratio a:b has a mean molecular weight 40. What would be mean molecular weight, if the gases are mixed in the ratio b:a under identical conditions? (gases are non-reacting):

The correct option is D 72Let mole fraction of O2 is x 40=32×x+80(1−x) Or x=56 a:b=x:(1−x)=56:16 When ratio is changed Mmixture=32×16+80×56=72

A mixture of $O_{2}$ and gas "Y" (mol.wt.80) in the mole ratio $a : b$ has a mean molecular weight 40. What would be mean molecular weight, if the gases are mixed in the ratio $b : a$ under identical conditions? (gases are non-reacting):

The correct option is D 72
Let mole fraction of $O_{2}$ is $x$
$40 = 32 \times x + 80 (1 - x)$
Or $x = \frac{5}{6}$
$a : b = x : (1 - x) = \frac{5}{6} : \frac{1}{6}$
When ratio is changed
$M_{m i x t u r e} = 32 \times \frac{1}{6} + 80 \times \frac{5}{6} = 72$