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Active mass of 5.6 lit N₂ at STP

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0.25M

22.4M

$\frac{1}{22 .4} M$

answer is C.

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Detailed Solution

Explanation

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.

Number of moles (n) = Volume of gas at STP / Molar volume at STP

n = 5.6 L / 22.4 L/mol

n = 0.25 mol

Active Mass = Number of moles / Volume (L)

Active Mass = 0.25 M

Since 0.25 is equivalent to 1/4 and can also be represented as 1/22.4 M when expressed as active mass in STP conditions.

Thus, the correct answer is: Option (c)

Final Answer:

Option (c): 1/22.4 M

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