What is the ratio of (a) gram atoms and (b) weights of the metals liberated during the electrolysis of fused sodium fluroide, magnesium fluoride and aluminium fluoride connected in a series?

Explanation

`Na^(+) + e^(-) to Na , Mg^(2+) + 2e^(-) to Mg and Al^(3+) + 3e^(-) to Al`
Ratio of gram atoms of metals liberated using one Faraday = `1/1 : 1/2 : 1/3 = 6 : 3 : 2`
Ratio of weights of metals liberated = `6 xx 23:3 xx 24:2 xx 27 = 23 : 12 : 9.`