A) \[44.8\text{ }L\]
B) \[11.2L\]
C) \[22.4\text{ }L\]
D) \[5.6\text{ }L\]
Correct Answer: D
Solution :
After passing same amount of electricity equal gram equivalent of substances deposited or dissolved. So, gram equivalent of aluminium = gram equivalent of hydrogen. \[\therefore \] \[\frac{weight\text{ }of\text{ }aluminium}{\begin{align} & weight\text{ }of\text{ }aluminium \\ & of\,\,\text{aluminium} \\ \end{align}}=\frac{volume\,\,of\,{{H}_{2}}\,at\,STP}{11.2L}\] \[\Rightarrow \] \[\frac{4.5}{9}=\frac{volume\,of\,{{H}_{2}}\,at\,STP}{11.2}\] \[\therefore \] Volume of \[{{H}_{2}}\] at STP \[=5.6\text{ }L\]
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