The mass of copper deposited from a solution of `CuSO_4` by passage of `45 A` current for `965` second is (Mol . Wt. pf Copper `= 63.5` ).A. ` 15. 875 g`B. ` 1. 585 g`C. ` 4825 g`D. ` 96 500 g`

The mass of copper deposited from a solution of `CuSO_4` by passage of

1.	The mass of copper deposited from a solution of `CuSO_4` by passage of `45 A` current for `965` second is (Mol . Wt. pf Copper `= 63.5` ).A. ` 15. 875 g`B. ` 1. 585 g`C. ` 4825 g`D. ` 96 500 g`
Answer» Correct Answer - B Current `I = 5 A` and time ` T = 965 sec` We know that equivalent weight of copper ` = ("Molecular weight")/("Valancy") = (63.5)/2` and quantitu of electricity passed in coulomb = current xx time ` =5 xxx 965 = 4825 C`. Since `96500` coulombs will deposit `(63.5)/2 g` of copper , therefore ` 4825` coulombs will deposit ` = (63. 5 xx 4825)/(96 500 xx 2) = 1. 5875 g`.

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The mass of copper deposited from a solution of `CuSO_4` by passage of `45 A` current for `965` second is (Mol . Wt. pf Copper `= 63.5` ).A. ` 15. 875 g`B. ` 1. 585 g`C. ` 4825 g`D. ` 96 500 g`

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