The active mass of 7.0 g of nitrogen in a 2.0 L container would be:

0.25

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0.125

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0.5

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Solution

The correct option is D 0.125
Moles of $N_{2} = \frac{7}{28}$

$= 0.25 m o l$

Active mass = $\frac{c o n c . i n m o l}{v o l u m e i n L}$

= $\frac{0.25}{2}$

= $0.125$

Hence, option $B$ is correct.

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The active mass of 7.0 g of nitrogen in a 2.0 L container would be:

The correct option is D 0.125Moles of N2=728 =0.25molActive mass = conc.in molvolume in L = 0.252 = 0.125Hence, option B is correct.

The correct option is D 0.125
Moles of $N_{2} = \frac{7}{28}$

$= 0.25 m o l$

Active mass = $\frac{c o n c . i n m o l}{v o l u m e i n L}$

= $\frac{0.25}{2}$

= $0.125$

Hence, option $B$ is correct.