The commercial aluminium (trivalent : atomic weight $27$ ) is generally obtained by electrolysis of $A 1 C 13$ . The charge required to deposit $13.5 g$ of aluminium is

0.5 F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.0 F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.5 F

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.0 F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $1.5 F$
Amount of aluminium present will be
$n u m b e r o f m o l e s = g i v e n w e i g h t / a t o m i c w e i g h t$
$n u m b e r o f m o l e s = 0.5$
Now 1 mole of aluminium requires 3 mole of electron or 3 Faraday(charge)
Therefore .5 mole will require 1.5 Faraday

Suggest Corrections

Similar questions

(

)

A l C l_{3}

54 g

= 108

= 27

= 27

54 g

Atomic mass of Ag = 108 g / m o l

Atomic mass of Al = 27 g / m o l

Aluminium is obtained by the electrolysis of its oxide, __.

Join BYJU'S Learning Program