Question

$16.8$ litre gas containing $H_2$ and $O_2$ is formed at $NTP$ on electrolysis of water. What should be the weight of electrolysed water?

• A. $5g$
• B. $9g$
• C. $10g$
• D. $12g$

Answer 1

The correct option is B $9g$

We know, electrolysis reaction of $H_{2}O$

$2H_{2}O\to 2H_2+O_2$

$H_2$ and $O_2$ are liberated in the ratio 2:1

Assume $H_2=2x$ litres and $O_2=x$ litres

$\therefore 3x=16.8$

Therefore $x=16.8/3=5.6$

volume of electrolysed $H_{2}O\left(g\right)=2\times 5.6=11.2$

No. of moles of $H_{2}O=11.2/22.4=0.5$

Weight of $H_{2}O$ electrolysed $=0.5\times 18g=9g$
Option B is correct