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Volume of Gases and Number of Moles

A gaseous mix...

Question

A gaseous mixture contains $C H_{4}$ and $C_{2} H_{6}$ in equimolar proportion. The weight of $1.12 l$ of this mixture at STP is:

A

2.3 g

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B

0.8 g

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C

1.15 g

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D

11.5 g

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Solution

The correct option is D $1.15 g$
At STP, 22.4 L of a gas $= 1$ mole.

So 1.12 L of mixture $= \frac{1.12}{22.4} = 0.05$ moles.

Out of this, 0.025 moles are $C H_{4}$ and 0.025 moles are $C_{2} H_{6}$ .

The molar masses of $C H_{4}$ and $C_{2} H_{4}$ are 16 g/mol and 30 g/mol respectively.

Mass of $C H_{4} = 16 g / m o l \times 0.025 m o l = 0.4$ g.

Mass of $C_{2} H_{6} = 30 g / m o l \times 0.025 m o l = 0.75$ g.

Total mass of the mixture $= 0.4 + 0.75 = 1.15$ g.

Hence, option $C$ is correct.

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