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

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

Delhi UMET/DPMTDelhi UMET/DPMT 2008Some Basic Concepts of Chemistry

Solution:

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 $=\frac{\text { mol. wt. of metal chloride }}{e q . \text { wt }, \text { of metal }+35.5}$
$\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
$=9 \times 2.66=23.9$