Salts of $A$ (atomic mass $15$ ), $B$ (atomic mass $27$ ) and $C$ (atomic mass $48$ ) were electrolyzed using the same amount of charge. It was found that when $4.5 g$ of $A$ was deposited, the mass of $B$ and $C$ deposited were $2.7 g$ and $9.6 g$ . The valencies of $A$ , $B$ and $C$ respectively.

1, 3 and 2

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3, 1 and 3

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2, 6 and 3

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2, 3 and 2

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Solution

The correct option is C $2, 6 and 3$
According to Faraday's second law,
$\frac{mass of A}{eq. mass of A} = \frac{mass of B}{eq. mass of B} = \frac{mass of C}{eq. mass of C}$
$\Rightarrow \frac{4.5}{15 / n_{1}} = \frac{2.7}{27 / n_{2}} = \frac{9.6}{48 / n_{3}} (H e r e n_{1}, n_{2}, n_{3} represents velency)$
$\Rightarrow 0.3 n_{1} = 0.1 n_{2} = 0.2 n_{3}$

$\Rightarrow n_{1} : n_{2} : n_{3} = 2 : 6 : 3$

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Salts of A (atomic mass 15), B (atomic mass 27) and C (atomic mass 48) were electrolyzed using the same amount of charge. It was found that when 4.5 g of A was deposited, the masses of B and C deposited were 2.7 g and 9.6 g. The valences of A, B and C were respectively

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Salts of A (atomic mass 15), B (atomic mass 27) and C (atomic mass 48) were electrolyzed using the same amount of charge. It was found that when 4.5 g of A was deposited, the mass of B and C deposited were 2.7 g and 9.6 g. The valencies of A, B and C respectively.

The correct option is C 2,6 and 3According to Faraday's second law, mass of Aeq. mass of A=mass of Beq. mass of B=mass of Ceq. mass of C ⇒4.515/n1=2.727/n2=9.648/n3 (Here n1,n2,n3 represents velency) ⇒0.3n1=0.1n2=0.2n3 ⇒n1:n2:n3=2:6:3

Salts of $A$ (atomic mass $15$ ), $B$ (atomic mass $27$ ) and $C$ (atomic mass $48$ ) were electrolyzed using the same amount of charge. It was found that when $4.5 g$ of $A$ was deposited, the mass of $B$ and $C$ deposited were $2.7 g$ and $9.6 g$ . The valencies of $A$ , $B$ and $C$ respectively.