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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answer is 72.

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Detailed Solution

Let mole fraction of $O_{2}$ is x
$40 = 32 \times x + 80 (1 - x) o r x = \frac{5}{6}$
$a : b = x : (1 - x) = \frac{5}{6} : \frac{1}{6}$
When ratio is changed
$M_{mixture} = 32 \times \frac{1}{6} + 80 \times \frac{5}{6} = 72$

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