A current of strength ` 2.5 A` was passed through ` CuSO_4` solution for 6 minute `264` seconds. The amount of copper deposited is (At. Of ` Cu = 63. 5, 1 F = 96500 C`) .

A current of strength 2.5 amp was passed through CuSO4 solution for 6 minutes 26 seconds.The amount of copper deposited is (at. wt. of C u=63.5) cdot(F=96500 C) (a)0.3175g (b)3.175g (c)0.635g (d)6.35g

A current of 2.6 ampere is passed through CuSO_4 solution for 6 minutes 20 seconds. The amount of Cu deposited is (At. wt. of Cu=63.5, Faraday = 96500 C)-

A current of 2.6 ampere was passed through CuSO_4 solution for 380 sec. The amount of Cu deposited is (at .wt. of Cu (63.5):

A current of 0.25 amp is passed through CuSO_(4) solution for 45 minutes. Calculate the mass of copper deposited on the cathode (At.wt of Cu = 63 .)

20 ampere current is flowing through CuSO_4 solution for 60 minutes. Find the amonut of cupper deposited. (At .wt. of Cu = 63.5)