Question

The equivalent weight of metal is 4.5 and the molecular weight of its chloride is 80, then the atomic weight of the metal is:
(A) 18
(B) 9
(C) 4.5
(D) 36

Answer 1

Hint: Equivalent weight of substance refers to the mass of the substance that reacts exactly with an arbitrarily fixed amount of another substance. Equivalent weight of a metal can be given as:
\[\text{Equivalent weight = }\dfrac{\text{atomic weight}}{\text{valency}}\]

Complete step by step solution:
According to the question, we have
Equivalent weight of metal, E = 4.5
The molecular weight of the metal chloride, M = 80
Let the molecular formula of the metal chloride be $MC{{l}_{n}}$.
We know that the atomic mass of chlorine = 35.5
Let us consider the atomic weight of the metal to be ‘X’.
Then, the molecular weight of the $MC{{l}_{n}}$ will be calculated as: (X + 35.5 $\times $ n)
Since the given molecular weight of $MC{{l}_{n}}$ = 80 g. Therefore, we can write
X + 35.5n = 80
As we already know that equivalent weight of the metal is given as
\[\text{Equivalent weight (E) =}\dfrac{\text{atomic weight (X)}}{\text{valency (n)}}\]
As the given equivalent weight of the metal = 4.5, its atomic weight will be equal to
Atomic weight (X) = equivalent weight (E) $\times $ valence (n)
X = 4.5 $\times $ n = 4.5n
Substituting the value of X = 4.5n in the above equation, we get
X + 35.5n = 80 = 4.5n + 35.5n = 80
40n = 80
n = 40.
Thus, we have a valence of the metal, n = 2. We can now find the atomic mass of the metal, X as
Atomic weight (X) = equivalent weight (E) $\times $ valence (n) = 4.5 $\times $ n = 4.5 $\times $2 = 9.
Therefore, the atomic mass of the metal is 9.

Hence, the correct option is (B).

Note: Note that the valency of the metal is its capacity to combine with chlorine atoms. It is equal to the number of electrons the metal loses to gain the stable noble gas configuration. Since valency, in this case, is 2, the formula of the metal chloride will be $MC{{l}_{2}}$.