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Salts of $A, B$ and $C$ were electrolysed under identical conditions using the same quantity of electricity. It was found that when $2.1 \mathrm{~g}$ of $A$ was deposited, the weights of $B$ and $C$ deposited were $2.7 \mathrm{~g}$ and $9.6 \mathrm{~g}$. If the atomic mass of $A, B$ and $C$ respectively are 7,27 and 64 respectively, then the valancies of $A, B$ and $C$ respectively are
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The correct answer is:
$1,3,2$
Using concept of Faraday law,
Now, $A^{n_1^{+}}+n_1 e^{-} \longrightarrow A$
Similarly, $B^{n_2^{+}}+n_2 e^{-} \longrightarrow B$
and, $\mathrm{C}^{n_3^{+}}+n_3 e^{-} \longrightarrow C$
Since, charge is same then equivalent weight is same,
$$
\begin{array}{ll}
\therefore & \frac{2.1}{\frac{7}{n_1}}=\frac{2.7}{\frac{27}{n_2}}=\frac{7.2}{\frac{48}{n_3}} \\
\therefore & \frac{3 \times n_1}{10}=\frac{1 \times n_2}{10}=\frac{3 \times n_3}{2 \times 10} \therefore n_1=\frac{n_2}{3}=\frac{n_3}{2} \\
\therefore & n_1=1, n_2=3 \text { and } n_3=2
\end{array}
$$
Hence, valencies of $A, B$ and $C$ are 1,3 and 2 .
Now, $A^{n_1^{+}}+n_1 e^{-} \longrightarrow A$
Similarly, $B^{n_2^{+}}+n_2 e^{-} \longrightarrow B$
and, $\mathrm{C}^{n_3^{+}}+n_3 e^{-} \longrightarrow C$
Since, charge is same then equivalent weight is same,
$$
\begin{array}{ll}
\therefore & \frac{2.1}{\frac{7}{n_1}}=\frac{2.7}{\frac{27}{n_2}}=\frac{7.2}{\frac{48}{n_3}} \\
\therefore & \frac{3 \times n_1}{10}=\frac{1 \times n_2}{10}=\frac{3 \times n_3}{2 \times 10} \therefore n_1=\frac{n_2}{3}=\frac{n_3}{2} \\
\therefore & n_1=1, n_2=3 \text { and } n_3=2
\end{array}
$$
Hence, valencies of $A, B$ and $C$ are 1,3 and 2 .
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