Question

During thermal dissociation of a gas, the vapour density:
A.Remains the same
B.Increases
C.Decreases
D.Increases in some cases and decreases in others

Answer 1

Hint:
This question is asked from the chapter of gaseous state. To solve this problem, you need to know the concept that vapour density is directly proportional to the molar mass.

Complete step by step answer:
We can start answering this problem by considering a gas molecule A having molar mass M is thermally dissociated into small two gas molecules B and C with the molar masses ${m_1}$ and ${m_2}$. That is,
$A(g) \to B(g) + C(g)$ $A(g) \to B(g) + C(g)$
It is obvious that the molar masses ${m_1}$ and ${m_2}$ are smaller than the molar mass of the parent gas atom M. Because, the molar mass of the parent gas molecule is splitted into the small gas molecules during the thermal dissociation. We know that the vapour density is directly proportional to the molar mass of the gas molecules. If the molar mass increases, the vapour density of the gas molecule also increases. Here, during the thermal dissociation, the molar mass decreases. Therefore, the vapour density of the gas will decrease. Therefore, the correct option for this question is (c).

Note:
You may need to know more about vapour density. Vapour density is density of a vapour, which forms in comparison with hydrogen. It can be explained as a mass of a particular volume of a substance divided by the mass of the same volume of hydrogen. So, vapour density is a unit less quantity.