The mass of copper deposited from a solution of CuSO4 by passage of 5 A current for 965 second is (Mol. wt. of Copper = 63.5)

Question

The mass of copper deposited from a solution of CuSO4 by passage of 5 A current for 965 second is (Mol. wt. of Copper = 63.5) (A) 15.875 g (B) 1.5875 g (C) 4825 g (D) 96500 g

Tardigrade Admin · Accepted Answer

Current (I) = 5 A and time (t) = 965 sec We know that equivalent weight of copper =(Molecular weight/Valancy)=(63.5/2) and quantity of electricity passed in coulomb = current × time = 5 × 965 = 4825 C. Since 96500 coulombs will deposit (63.5/2) g of copper, therefore 4825 coulombs will deposit (63.5 ×4825/96500×2)=1.5875 g

Q. The mass of copper deposited from a solution of $C u S O_{4}$ by passage of $5 A$ current for $965$ second is (Mol. wt. of Copper = $63.5$ )

$15.875$ g

$1.5875$ g

$4825$ g

$96500$ g

Q. The mass of copper deposited from a solution of CuSO4​ by passage of 5A current for 965 second is (Mol. wt. of Copper = 63.5)

15.875 g

1.5875 g

4825 g

96500 g

Q. The mass of copper deposited from a solution of $C u S O_{4}$ by passage of $5 A$ current for $965$ second is (Mol. wt. of Copper = $63.5$ )

$15.875$ g

$1.5875$ g

$4825$ g

$96500$ g