A) 30 mL
B) 3 L
C) 0.3 L
D) 1.5 L
Correct Answer: B
Solution :
Given,\[R=0.0821\text{ }L\]atom mol\[{{s}^{-1}}{{K}^{-1}}\] \[T=27{}^\circ C=27+273=300K\] \[p=0.821\text{ }atm\] From \[pV=RT\] \[V=\frac{RT}{p}=\frac{0.0821\times 300}{0.821}\] or \[V=30\text{ }L\] Molecular weight of\[CO=12+16=28\] \[\because \]28 g of CO occupy volume = 30 L \[\therefore \]2.8 g of CO occupy volume\[=\frac{30}{28}\times 2.8\]
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