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Question: An element forms two oxides containing respectively. 50% and 40% by weight of the element. These oxi...

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

Explanation

Solution

Let us assume two oxide one is XOXO where XX is metal OO is oxide and another oxide could be XOMX{O_M} .
-Where Oxygen and Metal are in a different ratio. Assuming the total weight to be 100100 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 100gram100gram which contains 50gramsX50gramsX and 50gramsO’50grams'O
For second oxide total is 100gram100gram which contains 40gramsX40gramsX and 60gramsO’60grams'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 50gramsX50gramsX , then how many grams of O'O' is required
The number of grams of O'O'=6040×50=23\dfrac{{60}}{{40}} \times 50 = \dfrac{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 11 ratio and they form 22 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.