Question

When an electric current is passed through acidulated water for 1930 sec, 1120 ml of $H_2$ is liberated at cathode at STP. The current passed in amp is:

• A. 0.05
• B. 0.5
• C. 5
• D. 50

Answer 1

Answer: (C)

5

1120 ml of $H_2$ is liberated at STP.

$\therefore$ 1.12 litres of $H_2$ is liberated.

Moles of $H_2$ liberated $=\frac{1.12}{22.4}=0.05$ moles.

1 mole of $H_2$ requires 2 moles of $e^{-}$.

0.05 mole of $H_2$ requires $=\frac{0.05\times 2}{1}=0.1$ mole $e^{-}$.

The charge on $0.1$ mole $e^{-}=96500\times 0.1$.

$⟹96500\times 0.1=I\times 1930$

Therefore, $I=$ $\frac{9650}{1930}$ $=5$ $amp$.

Hence, option $C$ is correct.