MCQ Detailed View

This question is a classical application of the concept of equivalent weight and molecular weight calculation in analytical chemistry.

The formula for equivalent weight of an element is:

Equivalent weight=Atomic weightValency\text{Equivalent weight} = \frac{\text{Atomic weight}}{\text{Valency}}Equivalent weight=ValencyAtomic weight

We are told that the metal is bivalent, meaning its valency is 2. The equivalent weight of this metal is given as 12. By rearranging the formula:

$$\text{Atomic weight of metal}=\text{Equivalent weight}\times \text{Valency}$$ $$\text{Atomic weight}=12\times 2=24$$

Thus, the metal has an atomic weight of 24.

Now, we need to calculate the molecular weight of its oxide. Since the metal is bivalent, it combines with oxygen (valency = 2) in a 1:1 ratio. Therefore, the formula of its oxide is MO.

Molecular weight of MO = Atomic weight of metal + Atomic weight of oxygen

$$=24+16=40$$

Hence, the molecular weight of the oxide is 40.

This type of calculation is important in chemistry because it shows how equivalent weights can be used to find unknown atomic or molecular weights. The method is especially useful in classical analytical chemistry when elements or compounds were identified through quantitative relationships rather than advanced spectroscopic methods.

To summarize:

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  Equivalent weight helps in linking experimental results with theoretical atomic weights.

  
  


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  By knowing valency and equivalent weight, the atomic weight of a metal can be easily determined.

  
  


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  Adding oxygen’s atomic weight gives the molecular weight of the oxide.