Salts of A (atomic weight 8),B (atomic weight | vedclass

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(C) According to Faraday's second law of electrolysis,the mass of different substances deposited by the same quantity of electricity is proportional t

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$1, 3$ and $2$

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(C) According to Faraday's second law of electrolysis,the mass of different substances deposited by the same quantity of electricity is proportional to their equivalent masses $(E)$.$W \propto E$$W = \frac{E 	imes Q}{F}$Since $Q$ and $F$ are constant,$W \propto E = \frac{	ext{Atomic mass}}{	ext{Valency}}$Let the valencies of $A, B, C$ be $n_A, n_B, n_C$ respectively.$W_A : W_B : W_C = \frac{8}{n_A} : \frac{18}{n_B} : \frac{50}{n_C}$$2.4 : 1.8 : 7.5 = \frac{8}{n_A} : \frac{18}{n_B} : \frac{50}{n_C}$For $A$: $\frac{8}{n_A} = 2.4 \implies n_A = \frac{8}{2.4} = 3.33$ (This approach assumes proportionality constant $k=1$ for simplicity,let's use ratios).$n_A : n_B : n_C = \frac{8}{2.4} : \frac{18}{1.8} : \frac{50}{7.5}$$n_A : n_B : n_C = 3.33 : 10 : 6.66$Dividing by $3.33$:$n_A : n_B : n_C = 1 : 3 : 2$Thus,the valencies are $1, 3$ and $2$.

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