Question

A current of $3A$ was passed for 1 hour through an electrolyte solution of $A_x{B}_y$ in water. If $2.977$ g of A (atomic wright $106.4$) was deposited at the cathode and $B$ was a monovalent ion, the formula of electrolyte was:

• A. $A{B}_2$
• B. $AB$
• C. $A{B}_3$
• D. $A{B}_4$

Answer 1

The correct option is D $A{B}_4$

$\frac{W}{E_w}=\frac{It}{96500}(A^{y+}+y{e}^{-}\to A)$
$\frac{2.977}{\frac{106.4}{y}}=\frac{3\times 1\times 60\times 60}{96500}$
$\therefore y=4\Rightarrow ElectrolyteisA{B}_4.$