If $1$ g each of CO and $C H_{4}$ are taken at STP, $C O$ gas will occupy a greater volume.

True

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False

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Solution

The correct option is B False
Molar mass of :
$C O = 28 g (1 g r a m o f C O = \frac{22.4}{28} = 0.8 L)$
$H_{2} O = 18 g (1 g r a m o f H_{2} O = \frac{22.4}{18} = 1.24 L)$
$C H_{4} = 16 g (1 g r a m o f C H_{4} = 1.4 L)$
$N O = 20 g (1 g r a m 0 f N O = 1.12 L)$
Hence the compound having greatest volume is $C H_{4}$ and the lowest volume is $C O$ .

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