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Thermodynamic Processes

A gaseous mix...

Question

A gaseous mixture enclosed in a vessel of volume $^{'} V^{'}$ consists of $2$ gram mole of gas $A$ and $4$ gram mole of gas $B$ at a certain temperature $T$ . The gram molecular weights of the gases $A$ and $B$ are $4$ and $16$ respectively. The molecular weight of the mixture will be

A

12 g

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B

6 g

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C

4 g

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D

32 g

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Solution

The correct option is A $12 g$
Given,
Molecular weight of gas $A, (M_{A}) = 4 g$
Molecular weight of gas $B, (M_{B}) = 16 g$
No. of moles of gas $A$ in mixture $(n_{A}) = 2$
No. of moles of gas $B$ in mixture $(n_{B}) = 4$
Molecular weight of the mixture will be given by
$M = \frac{n_{A} M_{A} + n_{B} M_{B}}{n_{A} + n_{B}} = \frac{2 (4) + 4 (16)}{2 + 4}$
$= 12 g$

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