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A curent of `3A` was passed for 1 hour through an electrolyte solution of `A_(x) B_(y)` in water. If `2.977g ` of `A(` atomic weight `106.4)` was deposited at cathode and `B` was a monovalent ion, the formula of electrolyte wasA. `AB_(2)`B. `AB`C. `AB_(3)`D. `AB_(4)` |
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Answer» Correct Answer - d `(W)/(E_(w))=(It)/(96500)(A^(y+)+ye^(_)rarr A)` `(2.977)/(106.4/y)=(3xx1xx60xx60)/(96500)` `:.y=4implies` Electrolyte is `AB_(4)`. |
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