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The volume occupied by 4 g of $H_{2}$ at STP is

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33.6

22.4

44.8

11.2

answer is C.

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Detailed Solution

$\begin{array}{l} 1 m o l e o f H_{2} \to 2 {g H}_{2} \to 22.4 lit volume \\ 4 {g H}_{2} \to x lit \\ 2 \times x = 4 \times 22.4 \\ x = \frac{4 \times 22.4}{2} \\ = 44.8 lit v o l u m e o c c u p i e d \end{array}$

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