An unknown main-group element forms a highly volatile chloride. The equivalent weight of the metal i — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
An unknown main-group element $M$ forms a highly volatile chloride. The equivalent weight of the metal $M$ is determined to be exactly $4$. If the measured vapour density of this volatile chloride gas is $59.25$, determine the exact valency of the element $M$.
💡 Solution & Explanation
Let the valency of the element $M$ be $v$. The chemical formula of the chloride will be $MCl_v$. The atomic weight of $M = \text{Equivalent Weight} \times \text{Valency} = 4 \times v = 4v$. The molecular weight of the chloride compound = $M_w = \text{Atomic Weight of M} + (v \times 35.5)$. $M_w = 4v + 35.5v = 39.5v$. We are given the Vapour Density (V.D.) = $59.25$. By definition, Molecular Weight = $2 \times V.D. = 2 \times 59.25 = 118.5\text{ g/mol}$. Equating the two molar mass expressions: $39.5v = 118.5 \Rightarrow v = \frac{118.5}{39.5} = 3$. The valency of the element is $3$.