Active mass of 5.6 lit $N_{2}$ at STP is:

22.4 M

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0.25 M

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\frac{1}{22.4} M

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4 M

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Solution

The correct option is C $0.25 M$
No of moles in 22.4 l of $N_{2}$ = 1 mol

No of moles in 5.6 L = $\frac{1}{22.4} \times 5.6$ = 0.25 M

Hence, the correct option is $B$

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