Question

How long (approximate) should water be electrolysed by passing through $100$ amperes current so that the oxygen released can completely burn $27.66g$ of diborane?
[Atomic weight of $B=10.8u$]

• A. $6.4$ hours
• B. $0.8$ hours
• C. $3.2$ hours
• D. $1.6$ hours

Answer 1

The correct option is C $3.2$ hours

$B_2{H}_6+3O_2⟶B_2{O}_3+3H_{2}O$

27.66 gm of $B_2{H}_6$ i.e 1 mole of $B_2{H}_6$ requires 3 moles of $O_2$ .

Now oxygen is produced by electrolysis of water.

$2H_{2}O\stackrel{4F}{⟶}2H_2+O_2$

1 mole of $O_2$ is produced by 4F charge therefore 3 mole of $O_2$ will produced by 12F charge

By applying ,Q= It

$12\times 96500=100\times t$

$t=\frac{12\times 96500}{100\times 3600}$ hours

$t=3.2$ hours

Hence option C is correct option.