Q. 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:
NTA AbhyasNTA Abhyas 2022
Solution:
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, $a=2\times E=2\times \left(\frac{29}{2}\right)=29$
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, $a=2\times E=2\times \left(\frac{29}{2}\right)=29$