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

2.3 g

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0.8 g

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1.15 g

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11.5 g

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Solution

The correct option is B $1.15 g$
$22.4 L$ of a gas at STP $=$ 1 mole
$∴$ 1.12 L of mixture $= \frac{1.12}{22.4} = 0.05 m o l e s$

Since the gaseous mixture of $C H_{4}$ and $C_{2} H_{6}$ is an equimolar one,
0.05 moles will have 0.025 moles of $C H_{4}$ and 0.025 moles of $C_{2} H_{6}$

Weight of 0.025 moles of $C H_{4} =$ $0.025 \times 16 = 0.4 g$
Weight of 0.025 moles of $C_{2} H_{6} =$ $0.025 \times 30 =$ $0.75 g$

$∴$ Total weight of mixture $= 0.4 + 0.75 = 1.15 g$

Hence, option C is correct.

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