Question

The equivalent weight of a metal is $9$ and vapour density of its chloride is $59.25$. The atomic weight of metal is

• A. 23.9
• B. 27.3
• C. 36.3
• D. 48.3

Answer 1

Answer: (A)

Given, equivalent weight of metal $=9$
Vapour density of metal chloride $=59.25$
$\therefore$ Molecular weight of metal chloride
$=2\times$ vapour density
$=2\times 59.25=118.50$
$\because$ Valency of metal $=\frac{\text{ mol. wt. of metal chloride }}{eq.\text{ wt },\text{ of metal }+35.5}$
$\therefore$ Valency of metal $=\frac{118.5}{9+35.5}=\frac{118.5}{44.5}$
$=2.66$
Therefore, atomic weight of the metal
$=$ equivalent weight $\times$ valency
$=9\times 2.66=23.9$