A) \[4.0\,c{{m}^{3}}\]
B) \[56\,c{{m}^{3}}\]
C) \[604\,c{{m}^{3}}\]
D) \[8.0\,c{{m}^{3}}\]
Correct Answer: B
Solution :
\[\frac{Wt.\text{ }of\text{ }Cu\text{ }deposited}{Wt.\text{ }of\text{ }{{H}_{2}}\text{ }produced}=\frac{Eq.\text{ }wt.\text{ }of\text{ }Cu}{Eq.\text{ }wt.\text{ }of\text{ }H}\] \[\frac{0.16}{wt.\,of\,{{H}_{2}}}=\frac{64/2}{1}=\frac{32}{1}\] 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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