One ampere of current is passed for 9650 seconds through molten $A l C l_{3}$ . What is the weight in grams of $A l$ deposited at cathode?
(Atomic weight of $A l = 27$ )

0.9

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

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

0.09

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

Open in App

Solution

The correct option is A 0.9
$i = 1 A$ , $t = 9650 s$
Oxidation number of $A l$ in $A l C l_{3} = + 3$

$w = \frac{E_{e q} i t}{96500} = \frac{27 \times 1 \times 9650}{3 \times 96500}$

$w = \frac{9}{10} = 0.9 g m$

Suggest Corrections

Similar questions

0.5 A

A l C l_{3}

0.5 A

A l C l_{3}

The atomic weight of Al is 27. When a current of 5 Faradays is passed through a solution of Al⁺³ions, the weight of Al deposited is

A l C l_{3}

C

t

1 L

2 M C u S O_{4}

C u = 63.5

m

C u

Join BYJU'S Learning Program