Question

When a quantity of electricity is passed through $CuS{O}_4$ solution, $0.16g$ of Copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of $H_2$ liberated at $STP$ will be [Given $At.Wt.Cu=64$]

• A. $4.0c{m}^3$
• B. $56c{m}^3$
• C. $604c{m}^3$
• D. $8.0c{m}^3$

Answer 1

Answer: (B)

$\frac{\text{ Wt. of Cu deposited }}{\text{ Wt. of }{H}_2\text{ produced }}=\frac{\text{ Eq. }\text{ wt. of }Cu}{\text{ Eq. wt. of }H}$

$\frac{0.16}{wt\text{ . of }{H}_2}=\frac{64/2}{1}=\frac{32}{1}$

wt. of $H_2=\frac{0.16}{32}=5\times 10^{-3}g$

Volume of $H_2$ liberated at STP

$=\frac{22400}{2}\times 5\times 10^{-3}cc=56cc$