Question

A current of 10 ampere is passed through a molten lcl 3 for 96.5 seconds calculate the mass of al deposited

Answer 1

Answer: 0.09 gram

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 10 A

t= time in seconds = 96.5 s

$Q=10A\times 96.5s=965C$

$AlCl_3\rightarrow Al^{3+}+3Cl^-$

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

$96500\times 3=289500Coloumb$ of electricity deposits 1 mole of Al

965C of electricity deposits =$\frac{1}{289500}\times 965=0.0033moles$ of Al.

Mass of aluminium deposited =$moles\times {\text {Molar mass}}=0.0033\times 27=0.09 gram$