Question

On passing 1.5F charge the number of moles of aluminium deposited at cathode are:
Molar mass od aluminium - 27 gm mol inverse

Answer 1

When you don't know the answer to an exam question, you have two options: acknowledge defeat straightaway and prepare yourself to take the test again

Answer 2

Answer: 0.5 mole

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes

t= time in seconds

1 mole of electrons carry charge= 96500C = 1Faraday(F)

$Al^{3+}+3e^{-1}\rightarrow Al$

$96500\times 3=3F$ of electricity deposits 1 mole of  Al.

1.5F of electricity deposits =$\frac{1}{3}\times 1.5=0.5mole$ of Al.