Question

two oxides of a meta contained respectively 92.59% and 96.15% of the metal. show those facts to agree with the law of multiple proportions.

Answer 1

Q97354

Solution:

The law of multiple proportions states that: If two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers. ---(1)

So, let us denote the unknown metal as $M$ and let it's two oxides have formulas $MO_x$ and $MO_y$ .

$\therefore MO_x$ and $MO_y$ contain 92.59% and 96.15% of the metal $M$ respectively. (given) ---(2)

Let us assume the atomic weight of the metal $M$ to be $'p'gms.$

Thus, molecular weight of $MO_x=(p+16x)gms.$

And molecular weight of $MO_y=(p+16y)gms.$

$M+ xO\to MO_x$

$\implies 1 mol$ or $'p'gms.$ of the metal reacts with $16x gms.$ of Oxygen. ---(3)

$M+ yO\to MO_y$

$\implies 1 mol$ or $'p'gms.$ of the metal reacts with $16y gms.$ of Oxygen. ---(4)

From (1), (2) and (3);

For the given facts to agree with the Law of Multiple Proportions, we need to show that for a fixed amount of the metal $('p' gms.)$ , the ratio of Oxygen masses in the two oxides is a whole number.

$i.e.$ $(16x)/(16y)\iff (x/y)$ is a whole number.

(2)$\implies$ $p/(p+16x)=92.59$ %

$\implies 1/(1+(16x/p))=0.9259$

$\implies 16x/p=(0.9259)^{-1}-1$ ---(5)

And,$\implies p/(p+16y)=96.15$ %

$\implies 1/(1+(16y/p))=0.9615$

$\implies 16y/p=(0.9615)^{-1}-1$ ---(6)

Dividing (5) by (6), we get;

$x/y=(0.0741*0.9615)/(0.9259*0.0385)$

$=2$ (Answer)

The ratio x/y is 2, which is a whole number.

Hence Proved.

Thus, the given facts agree with the Law of Multiple Proportions.