Question

On passing $3A$ of electricity for $50$ min $1.8g$ metal deposits. The equivalent mass of metal is

• A. 9.3
• B. 19.3
• C. 38.3
• D. 39.5

Answer 1

Answer: (B)

$W_{\text{metal}}=\frac{\text{Eit}}{96500}$
$=\frac{E\times 3\times 50\times 60}{96500}$
$E=\frac{96500\times w}{3\times 50\times 60}$
$E=\frac{96500\times 1.8}{3\times 50\times 60}=19.3$