Question

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

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

Answer 1

The correct option is A 23.9 g

Let us assume the formula of the metal chloride to be $MC{l}_x$ ( x being the valency of the given metal )
The molar mass of metal chloride $=V.D\times 2$ = $59.25\times 2$ = 118.5 g.
$E_{\text{Metal choride}}=E_{\text{Metal}}+E_{\text{chloride ion}}$
$\frac{(\text{Molar mass}{)}_{MC{l}_x}}{x}=9+\frac{35.5}{1}$
$\frac{118.5}{x}$ = 9 + 35.5
$\Rightarrow$ x = 2.66
Hence $E_{\text{Metal}}=$ $\frac{(\text{Atomic weight}{)}_{\text{Metal}}}{(\text{Valency}{)}_{\text{Metal}}}=\frac{\text{Atomic weight}}{2.66}$

Atomic weight = 2.66 $\times$ 9 = 23.9 g