Question

One mole of chlorine combines with a certain weight of metal giving $111\,g$ of its chloride with the formula $MC{l_2}$ . The same amount of metal can displace $2\,g$ hydrogen from an acid. Find the equivalent weight of the metal.
A) $40$
B) $20$
C) $80$
D) $10$

Answer 1

Calculate the amount of chlorine present in one mole. From this we can calculate the amount of chlorine that reacted with the unknown metal. The remaining weight would give us the weight of the metal. From this we can calculate the amount of metal that can displace $1\,g$ of hydrogen.

Complete answer:
First, we need to calculate the amount of chlorine present in one mole.
$Number \,of\, moles = \dfrac{{weight}}{{mol.weight}}$
Given to us, number of moles $= 1$
Molecular weight of Chlorine molecule is $71\,g$ and therefore the weight of chlorine reacted would be $71\,g$
Now, we can calculate the weight of the metal as $111 - 71\,g$ which is $40\,g$
It is given to us that this amount of metal can displace $2\,g$ of hydrogen in an acid.
$40\,g$ of metal displaces $2\,g$ of Hydrogen, $1\,g$ of Hydrogen would be displaced by $20\,g$ of metal.

**Therefore the equivalent weight of the metal is $20\,g$ i.e. option B.

Note:**
In order to calculate the amount of Metal present in the given $111\,g$ of its chlorine, we find the amount of chlorine combined with the metal. This weight is removed from the metal chloride weight to acquire the weight of the metal alone. In order to find the equivalent weight of the metal, we calculate the amount of metal required to displace $1\,g$ of Hydrogen from the acid.