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

Options

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

Answer

C

Explanation

`B_(2) H_(6) + 3 O_(2) to B_(2) O_(3) + 2 H_(2)O`
Moles of `B_(2) H_(6) = (27.66)/(27.6) = 1` mol
1 mol of `B_(2) H_(6)` requires 3 mole of `O_2` for complete burning
Electrolysis of `H_2O`
` 2H_(2)O to O_(2) + 2 H_(2)`
1 mole of `O_2` is produced from 2 mole of `H_(2)O`
Amount of `H_2O` required to produce 3 mole of `O_(2)` = 6 mole
No. of Faradays required = `6 xx 2 = 12 F`
`12 xx 96500 C = I xx t`
`12 xx 96500 C = 100 xx t`
or `t = (12 xx 96500)/(100) = 11580 s`
or `(11580)/(3600) = 3.21` hr