When 0.25 ampere current was passed through a metal salt solution for 10 minutes, 0.1677 g of the metal was deposited. If specific heat of the metal is 0.0556 cal g-(degree)-, find the at. wt. of metal. (1 faraday = 96500 coulomb)

(i) Current passed, C = 0.25 ampere; time (t) in seconds = 10 x 60 = 600 s, Wt. of metal deposited = 0.1677 g∴g. eq. wt. of metal  = Wt. of metal deposited ×96500 coulomb  Quantity of electricity passed (=ct)=0.1677g×96500 coulomb (0.25 ampere ×600s) i.e., 150 coulomb =107.89g.(ii)  Approx. at. wt. =6.4 sp. heat =6.40.0556=115.1∴ Valency = Approx. at. wt.  Eq. wt. =115.1107.89≃1∴  Exact at. wt. = Eq. wt. × Valency =107.89×1=107.89