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Standard XII

Chemistry

Faraday's Second Law

An element A ...

Question

An element A (atomic mass 112) and element B (atomic mass 27) form chlorides. Solutions of these chlorides are electroysed separately and it is found that when the same quantity of electricity is passed, 5.6 g of A was deposited while in other cell only 0.9 g of B was depoosited. What is valence of A if valence of B is 3?

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Solution

Let the valency of A be x.
$\frac{mass of A \times valence of A}{Molar mass of A} = \frac{mass of B \times valence of B}{Molar mass of B}$
Then , $\frac{5.6 \times x}{112} = \frac{0.9 \times 3}{27}$
Hence, $x = 2$ .
Thus, the valence of A is $2.$

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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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Let the valency of A be x.mass of A×valence of AMolar mass of A=mass of B×valence of BMolar mass of BThen ,5.6×x112=0.9×327Hence, x=2.Thus, the valence of A is 2.

Let the valency of A be x.
$\frac{mass of A \times valence of A}{Molar mass of A} = \frac{mass of B \times valence of B}{Molar mass of B}$
Then , $\frac{5.6 \times x}{112} = \frac{0.9 \times 3}{27}$
Hence, $x = 2$ .
Thus, the valence of A is $2.$