Hydrogen peroxide and water contain $5.93\% $ and $11.2\% $ of hydrogen respectively. The data illustrates the law of?
A. Constant proportions
B. Multiple proportions
C. Reciprocal proportions
D. Conservation of mass

Hint: As we know that there are four laws of chemical combinations and one of them states that when two elements combine with each other to form two or more than two compounds, the masses of one of the elements which combines with fixed mass of other bears a simple whole number ratio.

Complete step by step solution:
Let us consider the given situations in the question:
For hydrogen peroxide, let us assume that $100g$ of sample is there which contains $5.93g$ of hydrogen, so the amount of oxygen in hydrogen peroxide will be $100 - 5.93 = 94.07g$. So, the ratio of mass of oxygen to the hydrogen in the given sample of hydrogen peroxide will be given as:
$\dfrac{{94.07}}{{5.93}} = 15.86$

Similarly, for water, let us assume that the $100g$ of sample contains $11.2g$ of hydrogen and the amount of oxygen in water will be $100 - 11.2 = 88.8g$. Now, we can calculate the ratio of mass of oxygen to the hydrogen in the given sample of water and it will be equivalent to:
$\dfrac{{88.8}}{{11.2}} = 7.93$
Thus, the two ratios are in the proportion $\dfrac{{15.86}}{{7.93}} = 2:1$.
Therefore, we can say that the given data illustrates the law of multiple proportions which states that when two elements react with each other to produce two or more than two compounds, then the masses of one of the elements that combine with the fixed mass of other bears a simple whole number ratio.

Hence, the correct answer is option (B).

Note: Law of constant proportion states that a chemical compound always possesses exactly the same proportion of elements by mass irrespective of the source. Law of conservation of mass states that during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactant. And lastly, the law of reciprocal proportion states that when two different chemicals react separately but with the same weight of a third chemical, then the ratio of masses is either the same or a simple multiple of the mass ratio in which they react.