Question

The chloride of a metal contains $71\%$ chlorine by weight and the vapour density of it is $50$ . The atomic weight 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 $\text{=}\frac{{\text{W}}_{\text{m}}}{{\text{E}}_{\text{m}}}\text{=}\frac{{\text{W}}_{\text{Cl}}}{{\text{E}}_{\text{Cl}}}\text{or}\frac{\text{29}}{\text{E}}\text{=}\frac{\text{71}}{\text{35}\text{.5}}$
$E=\frac{29}{2}$
Now $a+35.5n=100$
or $n\times E+35.5\times n=100$
$\therefore n=2$
Therefore, $a=2\times E=2\times \left(\frac{29}{2}\right)=29$