A mixture of $\mathrm{O}_{2}$ and gas "Y" mol. mass 80 in the mole ratio a $: \mathrm{b}$ has a mean…
A mixture of $\mathrm{O}_{2}$ and gas "Y" mol. mass 80 in the mole ratio a $: \mathrm{b}$ has a mean molecular mass 40. What would be mean molecular mass, if the gases are mixed in the ratio $\mathrm{b}:$ a under, identical conditions? (Assume that gases are non-reacting):
40
48
62
72
Solution
Let mole fraction of $\mathrm{O}_{2}$ is $\mathrm{x}$ $40=32 \times x+80(1-x)$
or $x=5 / 6$
$a: b=x:(1-x)=\frac{5}{6}: \frac{1}{6}$
When ratio is changed
$\mathrm{M}_{\text {mixture }}=32 \times \frac{1}{60}+80 \times \frac{5}{6}=72$