Average molar mass of a mixture of $C H_{4} a n d C_{2} H_{4}$ 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?

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answer is 24.

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Detailed Solution

Let a mole of $C H_{4}$ and b mole of $C_{2} H_{4}$ be present
Average molar mass of mixture $= \frac{a \times 16 + b \times 28}{a + b} = 20$
$∴ 4 a = 8 b o r \frac{a}{b} = \frac{1}{2} o r 2 a = b$
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 $= \frac{b \times 16 + a \times 28}{a + b} = 20$
$= \frac{b \times 16 + 2 b \times 28}{b + 2 b} = \frac{72}{3} (N o w, \frac{b}{a} = \frac{1}{2}) = 24 g m o l^{- 1}$