Question

Vapour density of metal chloride is 66. Its oxide contains 53% metal. What is the atomic weight of the metal?
A.21
B.54
C.27.06
D.2.086

Answer 1

Hint:The weight of metal oxide can be assumed to be 100 g. Thus, calculating the equivalent weight of metal. Which in turn can be used for calculating the valency of the metal. Then both of these will be used for atomic weight calculation since atomic weight is equal to equivalent weight $ \times $ valency.

Complete step by step answer:
Now, we will start solving the question step by step:
Step 1. Calculation of Equivalent weight of metal:
Let us assume the weight of metal oxide to be 100 g. Which means the weight of metal is 53 g and the weight of oxygen is 47 g.
Now, as we all know Equivalent weight of metal = $\dfrac{{{{\text{w}}_{\text{2}}}}}{{{{\text{w}}_{\text{1}}}}} \times {{\text{a}}_{\text{1}}}$, where, ${{\text{w}}_{\text{2}}}$ = weight of metal, ${{\text{w}}_{\text{1}}}$ = weight of oxygen and ${{\text{a}}_{\text{1}}}$ =atomic weight of oxygen
After putting all the given values we have equivalent weight of metal is = $\dfrac{{53}}{{47}} \times 8$ =9.02

Step 2. Calculation of Valency:
As we know, Valency = $\dfrac{{{\text{2}} \times {\text{VD}}}}{{{\text{E + 35}}{\text{.5}}}}$ where, VD is vapour density, E is equivalent weight. We can now put all the given values, Valency = $\dfrac{{{\text{2}} \times {\text{66}}}}{{{\text{9}}{\text{.02 + 35}}{\text{.5}}}}$
So, Valency =2.96. Since valency is always an integer, we will take Valency as 3

Step 3. Calculation of atomic weight of metal:
Atomic weight =Equivalent weight $ \times $ Valency =9.02 $ \times $ 3 =27.06
In the end we can conclude that the atomic weight of the metal is equal to 27.06.
Hence option (C) is the correct answer.

Note:
Always remember that valency of an element is always an integer, never in fractions. And due to this reason the valency which came out to be 2.96 is taken as 3 in the present question. One more thing to be remembered here is that while calculating the equivalent weight of species a1 metal in this case atomic weight of counter species i.e. a2, here oxygen is being used and vice-versa.