Average molar mass of a mixture of $C{H}_{4}and{C}_2{H}_4$  present in the mole ratio a:b is 20 g/mol. If the mole ratio is reversed (b:a),  what will be molar mass of mixture?

Let a mole of $C{H}_4$  and b mole of $C_2{H}_4$  be present
  Average molar mass of mixture  $=\frac{a\times 16+b\times 28}{a+b}=20$
 $$\begin{gathered} \therefore 4a=8b \\ or\frac{a}{b}=\frac{1}{2} \\ or2a=b \end{gathered}$$
 On reversing mole ratio b mole of $C{H}_4$ and a mole of $C_2{H}_4$ be present
 Average molar mass of mixture  $=\frac{b\times 16+a\times 28}{a+b}=20$
      $$\begin{gathered} =\frac{b\times 16+2b\times 28}{b+2b}=\frac{72}{3} \\ (Now,\frac{b}{a}=\frac{1}{2}) \\ =24gmo{l}^{-1} \end{gathered}$$