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Question: In a metal chloride, the weight of metal and chlorine is in the ratio of 1:2. The equivalent weight ...

In a metal chloride, the weight of metal and chlorine is in the ratio of 1:2. The equivalent weight of the metal will be:
(A) 71
(B) 35.5
(C) 106.5
(D) 17.75

Explanation

Solution

To solve this question, we first need to know what is equivalent weight. The mass of a substance that can combine or displace a fixed quantity of another substance is known as the equivalent weight of that substance.

Complete answer:
It is given to us that
weight of metalweight of chlorine=12\dfrac{\text{weight of metal}}{\text{weight of chlorine}}=\dfrac{1}{2}
Now, the equivalent weight of an element can be given by the mass of the element required to combine or displace either 1.008 g of H, 8.0 g of O, or 35.5 g of Cl.
The equivalent weight of an element can be determined by dividing the atomic weight by the valency of the elements.
Equivalent Weight = Atomic weight of elementValency of element\text{Equivalent Weight = }\dfrac{\text{Atomic weight of element}}{\text{Valency of element}}
Now since the valency of the Chlorine atom is 1, it combines with one atom of another element having valency 1 to form an ionic bond. So, the relationship between the equivalent weight and weight of metal and chlorine can be given by
Eq weight of metalweight of metal=Eq weight of chlorineweight of chlorine\dfrac{\text{Eq weight of metal}}{\text{weight of metal}}=\dfrac{\text{Eq weight of chlorine}}{\text{weight of chlorine}}
So,
Eq weight of metalEq weight of chlorine=weight of metalweight of chlorine=12\dfrac{\text{Eq weight of metal}}{\text{Eq weight of chlorine}}=\dfrac{\text{weight of metal}}{\text{weight of chlorine}}=\dfrac{1}{2}
Now, we know that the equivalent weight of chlorine is 35.5 g/eq.
So,

& \dfrac{\text{Eq weight of metal}}{35.5}=\dfrac{1}{2} \\\ & \text{Eq weight of meta}l=17.75\text{ g/eq} \\\ \end{aligned}$$ **Hence the equivalent weight of the metal will be option (D) 17.75 g/eq.** **Additional information:** Equivalent weight of a substance can be used to calculate the normality of a solution. $$N=\dfrac{w}{eq.w\times V}$$ Where N is the normality of the solution w is the weight of the solute eq.w is the equivalent weight of the solute V is the volume of the solution (in liters) **Note:** It should be noted that in acid-base reactions, the equivalent weight of the acid or the base depends on the moles of ${{H}^{+}}$ ions or $O{{H}^{-}}$ ions which are donated or reacted by it. Whereas in redox reactions, the equivalent weight of the reactant depends upon the moles of electrons that are donated or reacted by them.