Question

The average molar mass of a mixture of methane and ethene present in the ratio a:b is found to be $20g{mol}^{-}$. if the ratio were reversed,what would be the molar mass of the mixture?

Answer 1

Solution:-

Given that the methane an ethane present in the mixture in the ratio $a:b$.

Average molar mass of mixture $=20g/mol$

$\frac{16a+30b}{a+b}=20$

$\Rightarrow 16a+30b=20a+20$

$\Rightarrow 10b=4a\Rightarrow \frac{a}{b}=\frac{5}{2}$

If the ratio is reversed, i.e., methane an ethane present in the mixture in ratio $b:a$.

$\frac{16b+30a}{a+b}$ $=\frac{16+30\frac{a}{b}}{1+\frac{a}{b}}=\frac{16+30\left(\frac{5}{2}\right)}{1+\frac{5}{2}}=\frac{16+75}{\frac{7}{2}}=26$

Hence if the ratio is reversed, the molar mass of the mixture will be $26g/mol$.