A) \[4.0c{{m}^{3}}\]
B) \[56c{{m}^{3}}\]
C) \[604c{{m}^{3}}\]
D) \[8.0c{{m}^{3}}\]
Correct Answer: B
Solution :
\[\frac{Wt.\,\,of\,\,Cu\,deposited}{Wt.\,\,of\,{{H}_{2}}\,produced}=\frac{Eq.\,wt.\,of\,Cu}{Eq.\,wt.\,\,of\,H}\] \[\frac{0.16}{wt.\,of\,{{H}_{2}}}=\frac{64/2}{1}=\frac{32}{1}\] wt. of \[{{H}_{2}}=\frac{0.16}{32}=5\times {{10}^{-3}}g\] Volume of \[{{H}_{2}}\] liberated at STP \[=\frac{22400}{2}\times 5\times {{10}^{-3}}cc\] \[=56cc\]
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