A) 22.4 L
B) 44.8 L
C) 2.24 L
D) 4.48 L
Correct Answer: C
Solution :
M.wt of \[C{{O}_{2}}\] = 12+16+16 = 44 Volume of 44 gm of \[C{{O}_{2}}\] at NTP = 22.4 litre 1 gm of \[C{{O}_{2}}\] at NTP = \[\frac{22.4}{44}\] \[4.4\,gm\] of \[C{{O}_{2}}\] at N.T.P \[\Rightarrow \frac{22.4}{44}\times 4.4\]litre \[=2.24\] litreYou need to login to perform this action.
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