Question

A current of 9.65 A is passed through an aqueous solution of $NaCl$ using suitable electrodes for 1000 sec. Given that 1 faraday equals 96500 coulombs the amount of $NaOH$ (mol wt. 40) formed on electrolysis is:

• A. $2.0$ g
• B. $8.0$ g
• C. $4.0$ g
• D. $1.0$ g

Answer 1

The correct option is C $4.0$ g

Given:
$i=9.65A$, $t=1000sec$ M.W.=40 g/mol
$W_{deposited}=Z\times i\times t=\frac{E}{96500}\times 9.65\times 1000=\frac{\left(\frac{40}{1}\right)}{96500}\times 9.65\times 1000=4g$