If the compressibility factor of $H_{2}$ is $1.8$ , then the volume of one mole of real $H_{2}$ at STP will be:

40.32 L

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22.42 L

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30.322 L

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12.44 L

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Solution

The correct option is A $40.32 L$
Compressibility factor is used as a correction factor to ideal behaviour.
$V_{r e a l} = Z V_{i d e a l}$ (Z = compressibility factor)
$V_{r e a l} = 1.8 \times 22.4 L$
$V_{r e a l} = 40.322 L$

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