The oxides of two metals X and Y are isomorphous. The metal X whose atomic weight is 52 forms a chloride whose vapour density is 79.
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The oxides of two metals $X$ and $Y$ are isomorphous. The metal $X$ whose atomic weight is 52 forms a chloride whose vapour density is 79. The oxide of $Y$ contains $47 \cdot 1 \%$ of exygen. Calculate atomic weight of $Y$.

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SOLUTION :

Last portion of the problem gives eq. wt. of $Y$ and also from volatile chloride method valency of metal $X$ is found out, hence also the valency of $Y$ as the oxides of $X$ and $Y$ are isomorphous.

Wt. of oxygen $=47 \cdot 1 g$ and wt. of $Y=(100-47 \cdot 1) g =52 \cdot 9 g$

$\therefore \quad$ Eq. wt. of $Y=\frac{52 \cdot 9}{47 \cdot 1} \times 8=8 \cdot 98$

Let the valency of metal $X$ be $n$, hence formula of chloride of $X=X C l_{n}$.

Its vapour density $($ given $)=79, \therefore$ Mol. wt. $=2 \times$ V.D. $=79 \times 2=158$

From formula, mol. wt. of $X C l_{n}=$ at. wt. $X+n \times 35 \cdot 5=52+35 \cdot 5 n$

$\therefore \quad 52+35 \cdot 5 n=158, \quad \therefore 35 \cdot 5 n=158-52=106, \quad \therefore \quad n=\frac{106}{35 \cdot 5} \approx 3$

$\therefore$ Also valency of $Y=3$.

Atomic wt. of $Y=$ eq. wt. $\times$ valency $=8 \cdot 98 \times 3=26 \cdot 94$.

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