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Active mass of 5.6 lit N2 at STP...

Active mass of 5.6 lit `N_2` at STP

A

22.4M

B

0.25M

C

`1/(22.4)M`

D

4M

Text Solution

Verified by Experts

The correct Answer is:
C
Active mass = `(n)/(v//"(lit)")=((5.6)/(22.4))/(5.6)=(1)/(22.4)M`
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