A mixture of $${O_2}$$  and gas $$''Y''$$  $$mol.$$  mass 80 in the mole ratio $$a : b$$  has a mean molecular mass 40. What would be mean molecular mass, if the gases are mixed in the ratio $$b : a$$  under, identical conditions? ( Assume that gases are non-reacting ) :

Solution :
$$\eqalign{ & {\text{Let mole fraction of}}\,{O_2}\,{\text{is}}\,x \cr & 40 = 32 \times x + 80\left( {1 - x} \right) \cr & {\text{or}}\,\,x = \frac{5}{6} \cr & a:b = x\,:\left( {1 - x} \right) = \frac{5}{6}:\frac{1}{6} \cr & {\text{When ratio is changed}} \cr & {M_{mixture}} = 32 \times \frac{1}{{6}} + 80 \times \frac{5}{6} = 72 \cr} $$