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Hardness of Water and Treatment

In a containe...

Question

In a container, 92 g of $N_{2} O_{4}$ are present initially which dissociates into $N O_{2}$ as per the reaction,
$N_{2} O_{4} \to 2 N O_{2}$
After some time, 23 g of $N O_{2}$ is present in the gaseous mixture. Calculate the average molar mass of the mixture?

A

36.8 g/mol

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B

73.6 g/mol

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C

147.2 g/mol

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D

85.4 g/mol

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Solution

The correct option is B 73.6 g/mol
For the reaction,
$N_{2} O_{4} \to 2 N O_{2}$
$Initial moles of N_{2} O_{4} = \frac{m a s s o f N_{2} O_{4}}{m o l a r m a s s o f N_{2} O_{4}} = \frac{92 g}{92 g / m o l} = 1 m o l$

$Moles of N O_{2} i n m i x t u r e = \frac{m a s s o f N O_{2}}{m o l a r m a s s o f N O_{2}} = \frac{23 g}{46 g / m o l} = 0.5 m o l$

According to stoichiometry,
2 moles of $N O_{2}$ are produced by the dissociation of 1 mol of $N_{2} O_{4}$
$∴$ 0.5 moles of $N O_{2}$ are produced by the dissociation of 0.25 moles of $N_{2} O_{4}$
Moles of $N_{2} O_{4}$ left in mixture = 1 - 0.25 = 0.75 moles
For a gaseous mixture having $n_{1}$ and $n_{2}$ moles & $M_{1}$ and $M_{2}$ molar masses of $N_{2} O_{4}$ and $N O_{2}$ respectively,
$\Rightarrow Average molar mass of the mixture = \frac{T o t a l m a s s o f m i x t u r e}{T o t a l m o l e s o f m i x t u r e} = \frac{M_{1} \times n_{1} + M_{2} \times n_{2}}{n_{1} + n_{2}} = \frac{92 \times 0.75 + 46 \times 0.5}{0.75 + 0.5} = 73.6 g / m o l$

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