Question

The chloride of a metal contains $71\%$ chlorine by weight and the vapour density of it is $50$. The atomic weigh of the metal will be:

• A. $29$
• B. $58$
• C. $35.5$
• D. $71$

Answer 1

Answer: (A)

Mol. wt. of metal chloride $=50\times 2=100$
Let metal chloride be metal chlorine then
Equivalent of metal = Equivalents of chlorine
$=\frac{W_m}{E_m}=\frac{W_{Cl}}{E_{Cl}}$ or $\frac{29}{E}=\frac{71}{35.5}$
$E=\frac{29}{2}$
Now $a+35.5n=100$
or $n\times E+35\times 5n=100$
$\therefore n=2$
Therefore, $a=2\times E=2\times \left(\frac{29}{2}\right)=29$