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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? |
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Answer» SOLUTION :`Na^(+) e^(-) rarr Na, Mg^(2+) +2e^(-) rarr Mg` and `Al^(3+)+3e^(-) rarr Al` Ratio of gram ATOMS of metals liberated using one Faraday `= 1/1:1/2:1/3=6:3:2`, `w_(1) :w_(2):w_(3)=E_(1):E_(2):E_(3)` Ratio of weights of metals liberated `=23:12:9` |
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