Question

Vapour density of a metal chloride is $66$. Its oxide contains $53\%$metal. The atomic weight of the metal is:
A.$21$
B.$54$
C.$27.06$
D.$2.086$

Answer 1

Vapour density: It is defined as the density of a gas or vapour relative to that of hydrogen at the same pressure and temperature.
We know that relation between the vapour density and molecular mass is as follows:
Molecular mass is twice as that of vapour density.

Complete step by step answer:
First of all we will talk about vapour density and formula which connects vapour density to molecular weight.
Vapour density: It is defined as the density of a gas or vapour relative to that of hydrogen at the same pressure and temperature. And the formula is as: Molecular mass is twice as that of vapour density.
Now, in this question we are given that vapour density is $66$.
Let’s assume that the weight of metal oxide is $100gm$
Then the weight of metal will be $53gm$ as in the question it is given that $53\%$ metal is present in the metal oxide. So the weight of oxygen will be $47gm$
Now, equivalent weight of metal $= \dfrac{{{\text{weight of metal}}}}{{{\text{weight of oxygen}}}} \times 8$ as we know that equivalent weight of any metal oxide is equal to the weight of metal divided by weight of oxygen multiplied by $8$.
$=$ $\dfrac{{53}}{{47}} \times 8 = 9.02$
Now, valency is equal to twice the vapour density divided by its empirical weight.
Valency$= \dfrac{{2 \times {\text{ vapour density}}}}{{{\text{molecular weight}}}}$equivalent weight in this question is $E + 35.5$where $E$ is for metal and $35.5$ is for chlorine. So valency $= \dfrac{{2 \times 66}}{{9.02 + 35.5}} = 2.96$ which is similar to $3$
So atomic weight $=$equivalent weight $\times$valency
Atomic weight $= 9.02 \times 3 = 27.06$.
So option C is correct.

Note:
Remember that we have to multiply equivalent weight by valency of the atom to get the molecular mass of the atom. In the formula of vapour density and molecular weight we have to use molecular weight instead of equivalent weight.