One mole of a mixture of $N_{2}, {NO}_{2}$ and $N_{2} O_{4}$ has a mean molar mass of $55.4$ . On heating to a temperature at which all the $N_{2} O_{4}$ may be presumed to have dissociated; $N_{2} O_{4} ⇌ 2 {NO}_{2}$ , the mean molar mass tends to the lower value of $39.6$ . What is the mole ratio of $N_{2} : {NO}_{2} : N_{2} O_{4}$ in the original mixture?

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$0.5 : 0.1 : 0.4$

$0.5 : 0.2 : 0.3$

$0.6 : 0.1 : 0.3$

$0.6 : 0.2 : 0.2$

answer is A.

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Detailed Solution

$\begin{array}{lll} N_{2} {NO}_{2} N_{2} O_{4} \\ (1 - a - b) mol a mol b mol \end{array}$

$\begin{array}{l} (1 - a - b) \times 28 + a \times 46 + b \times 92 = 55.4 \\ N_{2} O_{4} ⇌ 2 {NO}_{2} \\ b a \\ \frac{(1 - a - b) \times 28 + (2 b + a) \times 46}{1 - a - b + 2 b + a} = 39.6 \\ (i) \div (ii) gives [1 + b] = \frac{55.4}{39.6} \\ \Rightarrow b = 0.4 \\ \Rightarrow a = 0.1 \end{array}$