Question

An element forms two oxides containing respectively. 50% and 40% by weight of the element. These oxides illustrate the:
A.Law of constant composition
B.Law of multiple proportion
C.Law of reciprocal proportion
D.None of these

Answer 1

Hint:Let us assume two oxide one is $XO$ where $X$ is metal $O$ is oxide and another oxide could be $X{O}_M$ .
-Where Oxygen and Metal are in a different ratio. Assuming the total weight to be $100$ since we have the percentage values in the question. we know that the weight of two elements will be equal to their percentage.

Complete step by step answer:
The law of constant proportions is also called the Proust’s law.
The ratio of elements in non-stoichiometric compounds varies from point to point. Samples of elements that change in their isotopic composition can also defy the law of definite proportions since the masses of two different isotopes of a component are different.
The law of reciprocal proportions is one of the most basic laws of stoichiometric chemistry. It relates the proportions in which elements combine across the various compounds.
Law of multiple proportions, statement that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in the ratio of whole numbers(small)
Here in this question given that fixed element so same metal is present in two oxide with fixed atomic weight
Let First oxide total is $100gram$ which contains $50gramsX$ and ${}^{\prime }50gram{s}^{\prime }O$
For second oxide total is $100gram$ which contains $40gramsX$ and ${}^{\prime }60gram{s}^{\prime }O$
Since the metal is the same in both the compounds, the mass of the metal in the compound will also be the same.
For second oxide: contains $50gramsX$ , then how many grams of ${}^{\prime }{O}^{\prime }$ is required
The number of grams of ${}^{\prime }{O}^{\prime }$=${\displaystyle \frac{60}{40}}\times 50={\displaystyle \frac{2}{3}}$
So the ratio is in a simple whole number. this property is suitable for the Law of multiple proportion
Hence correct answer to this question is option “B”.

Note:
When Atoms of the same element combine in more than $1$ ratio and they form $2$ or more compounds. The Ratios of their weight must be in a fixed ratio since Same metal is present which has a fixed atomic number.