Question

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

Answer 1

Hint: In this question, we will be seeing the valencies of the elements, how can one find the valency of one element separately, how chlorides are electrolysed and the element decomposition by passing electricity.

Complete step by step solution:
As we need to find the valency of A we know that it is ‘x’.
Then, $\dfrac{{\text{mass of A} \times \text{valency of A}}}{{\text{Molar mass of A}}}$ $ = $ $\dfrac{{\text{mass of B} \times \text{valency of B}}}{{\text{Molar Mass of B}}}$
Now, Let’s substitute all the values given to us in the question in the above equation,
Then, $\dfrac{{5.6 \times x}}{{112}}$ $ = $ $\dfrac{{0.9 \times 3}}{{27}}$
Solving the above substituted equation,
$x = 2$

Thus, the valency of A element is 2.

Additional information:
What is a Valency?
In chemistry, the valence or valency of a component is a proportion of its consolidating power with different particles when it structures chemical compounds or atoms.
What is Molar Mass?
In chemistry, the molar mass of a chemical compound is characterized as the mass of a sample of that compound separated by the measure of substance in that example, estimated in moles. The molar mass is an intensive property of the substance that doesn't rely upon the size of the sample. In the International System of Units (SI), the base unit of molar mass is kg/mol. However, molar masses are quite often communicated in g/mol. The molar mass of molecules of an element is given by the overall relative atomic mass of the element multiplied by the molar mass constant. The molar mass of a compound is given by the sum of the relative atomic mass ${A_r}$ of the atoms which form the compound multiplied by the molar mass constant ${M_u}$:
$M = {M_u}{M_r} = {M_u}\sum\limits_i {{A_{ri}}} $

Note: You need to remember the formulae and also you need to remember the molar masses of the elements so that you can substitute them in the formula to get the solution.