When a quantity of electricity is passed through CuSO_4 solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of H_2 liber…

Question

When a quantity of electricity is passed through CuSO_4 solution, 0.16 g 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. of Cu=64 ]

Accepted Answer

Correct answer: B. (b) : wt . of Cu deposited wt . ofH_2 produced= eq . wt . of Cu eq . wt . ofH or, 0.16 wt . ofH_2= 321 = 64/21 or, wt. of H_2= 0.1632=5 10^-3 g ∴ Volume of H_2 liberated at STP = 224002 5 10^-3 cc=56 cc ,

$When a quantity of electricity is passed through CuSO_{4} solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of H_{2} liberated at STP will be [given : at. wt. of Cu = 64]$

Solution

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