Salts of A, B and C were electrolysed under i | vedclass

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(B) According to Faraday's second law of electrolysis,the mass of a substance deposited is proportional to its equivalent weight $(E)$.$\frac{W_A}{E_A

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$1, 3, 2$

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(B) According to Faraday's second law of electrolysis,the mass of a substance deposited is proportional to its equivalent weight $(E)$.$\frac{W_A}{E_A} = \frac{W_B}{E_B} = \frac{W_C}{E_C}$Since $E = \frac{	ext{Atomic mass (M)}}{	ext{Valency (v)}}$,we have:$\frac{W_A 	imes v_A}{M_A} = \frac{W_B 	imes v_B}{M_B} = \frac{W_C 	imes v_C}{M_C}$Substituting the given values:$\frac{2.1 	imes v_A}{7} = \frac{2.7 	imes v_B}{27} = \frac{9.6 	imes v_C}{64}$$0.3 v_A = 0.1 v_B = 0.15 v_C$Dividing by $0.05$,we get $6 v_A = 2 v_B = 3 v_C$.For $v_A = 1, v_B = 3, v_C = 2$,the equation holds true:$0.3(1) = 0.1(3) = 0.15(2) = 0.3$.Thus,the valencies are $1, 3, 2$.

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