The vapour density of a highly volatile metal chloride is . Analytical data shows that its correspon — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
The vapour density of a highly volatile metal chloride is $66$. Analytical data shows that its corresponding metal oxide contains exactly $53\%$ of the metal by mass. Determine the exact atomic weight of the metal.
💡 Solution & Explanation
Step 1: Calculate Equivalent Weight (E) of the metal from the oxide. % Metal = 53%, % Oxygen = 47%. $E = \frac{\text{Mass of Metal}}{\text{Mass of Oxygen}} \times 8 = \frac{53}{47} \times 8 = 9.02\text{ g/eq}$. Step 2: Relate Molar Mass of chloride to its Valency (v). Molar Mass of Chloride = $2 \times V.D. = 2 \times 66 = 132\text{ g/mol}$. The formula is $MCl_v$. So, $Atomic Weight (A) + 35.5v = 132$. Step 3: Substitute $A = E \times v$. $9.02v + 35.5v = 132 \implies 44.52v = 132 \implies v \approx 3$. Step 4: Calculate Atomic Weight. $A = E \times v = 9.02 \times 3 = 27.06 \approx 27$. The metal is Aluminium.