A) 1 L
B) 11.2 L
C) 22.4 L
D) 4 L
Correct Answer: A
Solution :
Key Idea: (i) Molecular weight \[=2\text{ }\times \]vapour density (ii) Molecular weight in gram \[=22.4\,L\] at STP Given vapour density \[=11.2\] Molecular weight \[=2\text{ }\times \text{ }11.2=22.4\] \[\therefore \] 22.4 g of gas occupies \[\text{=}\,\text{22}\text{.4 L}\]at STP \[\therefore \]1 g of gas occupies \[=\frac{22.4}{22.4}\times 1\] \[=1\,L\]at STP
You need to login to perform this action.
You will be redirected in
3 sec
