On passing a current through molten KCl,19.5 | vedclass

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(A) According to Faraday's second law of electrolysis,the mass of substances deposited by the same quantity of electricity is proportional to their eq

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(A) According to Faraday's second law of electrolysis,the mass of substances deposited by the same quantity of electricity is proportional to their equivalent weights: $\frac{W_{K}}{E_{K}} = \frac{W_{Al}}{E_{Al}}$.For potassium $(K^+)$,the equivalent weight $E_{K} = \frac{39}{1} = 39 \ g/eq$.For aluminium $(Al^{3+})$,the equivalent weight $E_{Al} = \frac{27}{3} = 9 \ g/eq$.Given $W_{K} = 19.5 \ g$,we have $\frac{19.5}{39} = \frac{W_{Al}}{9}$.$W_{Al} = \frac{19.5 	imes 9}{39} = 0.5 	imes 9 = 4.5 \ g$.

On passing a current through molten $KCl$,$19.5 \ g$ of potassium is deposited. If the same quantity of electricity is passed through molten aluminium chloride,then the amount of aluminium deposited will be ............. $g$.

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