Question

The vapour density of a chloride of an element is 39.5. The equivalent weight of the element is 3.82. The atomic weight of the element is:
[A] 15.28
[B] 7.64
[C] 3.82
[D] 11.46

Answer 1

Hint: To solve this, firstly find out the molecular weight from the vapour density. Then assume that the metal chloride is $MC{{l}_{x}}$ and use this to find its molecular weight and equate it with the obtained molecular weight and solve for x. Use the relation that equivalent weight is molecular weight divided by the valency factor to get the answer.

Complete step by step answer:
We know that vapour density of a gas is the density of that gas with respect to that of hydrogen at the same temperature and pressure.
We can define vapour density with respect to molecular weight too. We know the molecular weight of a substance is the mass of one mole of the substance divided by the mass of one hydrogen atom. Similarly, vapour density is the mass of a certain volume of gas divided by the same volume of hydrogen.
Therefore, we can write the relation that molecular weight is equal to twice the vapour density.
Here, the vapour density of a metal chloride is given to us. Let us consider that the metal chloride is $MC{{l}_{x}}$ . Therefore, we can write the molecular weight of the metal chloride as – Atomic weight of metal + ‘x’ times the atomic weight of chlorine (as there are ‘x’ chlorine atoms).
But from the definition of vapour density we have that molecular weight is twice that of vapour density and vapour density is given to us that is 39.5.
Therefore, molecular weight will be $2\times 39.5=79$
So, we can write that-
     \[M+\left( x\times 35.5 \right)=79\]
The atomic weight of chlorine is 35.5.

Now, we know that equivalent weight is equal to the molecular weight upon valency factor.
The equivalent weight of the element ‘M’ is given to us as 3.82.
Valency factor is the number of electrons an element can donate or gain. Here, we have ‘x’ chlorine atoms. Therefore, the valency factor of the element will be x.
Therefore, $Equivalent\text{ weight=}\dfrac{Molecular\text{ weight }}{x}$
Or, $Molecular\text{ weight = equivalent weight}\times \text{x=3}\text{.82}\times \text{x}$
Now, if we put this in the above equation of molecular weight, we will get-
     \[\begin{align}
 & \left( 3.82\times x \right)+\left( x\times 35.5 \right)=79 \\
 & or,x\left( 35.5+3.82 \right)=79 \\
 & or,x=\dfrac{79}{39.32}=2.0091\simeq 2 \\
\end{align}\]

Therefore, we got x i.e. the valency factor from the above calculation as 2.
Now, we have to find the atomic weight of the element, which is x times of the equivalent weight. We have found out that x is 2. Therefore, it will be twice of equivalent weight.
Therefore, $Atomic\text{ weight = equivalent weight}\times \text{x=3}\text{.82}\times \text{2=7}\text{.64}$ .
So we can see from the above calculation that the atomic weight of the element is 7.64.
So, the correct answer is “Option B”.

Note: We should not be confused between equivalent weight and number of equivalents. Equivalent weight is the gram molecular weight of the substance divided by the valency factor of the substance. Valency factor is basically the charge it carries whereas the number of equivalents is the weight/ mass of the compound divided by its equivalent weight.