Question

The equivalent weight of metal is triple to that of oxygen. What is the ratio of weight of its oxide to that of metal?
A) 1 : 3
B) 1 : 3
C) 4 : 3
D) 3 : 4

Answer 1

The mass of one equivalent, that is, the mass of a given material that will mix with or displace a specified quantity of another substance, is known as equivalent weight. The mass of an element that combines with or displaces 1.008 grams of hydrogen, 8.0 grams of oxygen, or 35.5 grams of chlorine is its equivalent weight. These figures are calculated by dividing the atomic weight by the typical valence.

Complete answer:
A chemical molecule with at least one oxygen atom and one additional element in its chemical formula is known as an oxide. The dianion of oxygen, "oxide," is an ${{O}^{2}}$ (molecular) ion. As a result, metal oxides usually include an oxygen anion in the oxidation state of - 2. The majority of the Earth's crust is made up of solid oxides, which are formed when elements are oxidised by oxygen in air or water.
To get this ratio, you don't need the oxide formula; all you need is the corresponding weight of metal. As you may know, the equivalent weight of a metal that reacts with oxygen to create metal oxides is the mass of the metal combined with 8 parts by mass of oxygen. Because the equivalent weight of oxygen in oxides is 8 g, you can calculate the equivalent weight of the metal in this oxide to be $3\text{ }\times \text{ }8\text{ }g\text{ }=\text{ }24\text{ }g$ . This indicates that every 24 g of metal will mix with 8 g of oxygen to produce this oxide. To put it another way, if you utilise 24 g of metal to make this oxide, you can be sure it will include 8 g of oxygen and have a total mass of $24\text{ }g\text{ }+\text{ }8\text{ }g\text{ }=\text{ }32\text{ }g$ .
$\dfrac{32g}{24g}=\dfrac{4}{3}$
Hence option C is correct.

Note:
Most metal oxides have polymeric structures. The oxide usually connects three metal atoms (e.g., rutile structure) or six metal atoms (e.g., rutile structure) (carborundum or rock salt structures). The solids are insoluble in solvents due to the strong M-O bonds and the fact that these compounds are crosslinked polymers, however they are attacked by acids and bases. Many of the formulae are deceptively easy, despite the fact that many of the compounds are nonstoichiometric.