A) 1.5L
B) 0.3L
C) 3L
D) 30L
Correct Answer: C
Solution :
We know that \[PV=nRT\] Number of moles of \[C{{O}_{2}}\] \[(n)=\frac{\text{Wt}\text{.of}\,C{{O}_{2}}}{\text{mol}\text{.Wt}\text{.of}\,\text{C}{{\text{O}}_{\text{2}}}}\] \[=\frac{2.8}{28}=0.1\] \[T=27+273=300K\] \[P=0.821\,atm\] \[R=0.0821\] Then, \[V=\frac{nRT}{P}=\frac{0.1\times 0.0821\times 300}{0.821}\] \[=3\,L\]You need to login to perform this action.
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