A) 0.5 litre
B) 1.51itrc
C) 3 litre
D) 6 litre
Correct Answer: C
Solution :
The volume of carbon monoxide\[=\frac{nRT}{P}\] Mass of carbon monoxide = 2.8 g \[P=0.821\text{ }atm,\text{ }T=27+273=300\text{ }K\] \[R=0.0821\] \[\because \]gram molecular weight of \[CO=12+16=28\] \[\therefore \]number of moles in 2.8 g of \[CO=\frac{2.8}{28}=0.1\] \[\therefore \]volume\[=\frac{0.1\times 0.0821\times 300}{0.821}=3\]litreYou need to login to perform this action.
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