Answer
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Hint: Valence is the number of electrons that an element can give or take or share. Valance is also the ability of the element to combine with other atoms or groups of atoms. The equivalent mass of any substance can be taken for 8g of oxygen. The reaction of any metal with oxygen is known as oxidation reaction.
Complete step by step solution:
It is given in the problem that the metal M and oxygen combine to form a molecular formula ${M_x}{O_y}$ and we need to find the atomic mass of metal M.
Let the atomic mass of the metal M be A, the mass of the metal M in grams is equal to $xA$ grams and the mass of oxygen is equal to $\dfrac{y}{2} \times 32$grams.
The reaction of the given problem is equal,
$ \Rightarrow \underbrace {xM}_{\left( {xA} \right)g} + \underbrace {\left( {\dfrac{y}{2}} \right){O_2}}_{\left( {\dfrac{y}{2} \times 32} \right)} \to {M_x}{O_y}$
The equivalent mass of oxygen is equal to mass of metal M is,
$ \Rightarrow \left( {\dfrac{y}{2} \times 32} \right)g{\text{ of }}{{\text{O}}_2} \to x \times \left( A \right)g{\text{ of M}}$
For 8g ${{\text{O}}_2}$ the equivalent mass of the metal M is equal to,
$ \Rightarrow E = 8g{\text{ of }}{{\text{O}}_2} \to \left( {\dfrac{{x \times A}}{{2y}}} \right)g{\text{ of M}}$
Here E is the equivalent mass.
$ \Rightarrow E = \left( {\dfrac{{x \times A}}{{2y}}} \right)g$
The atomic mass of the metal M is equal to,
$ \Rightarrow A = \dfrac{{2Ey}}{x}$.
The atomic mass of the metal M is equal to $\dfrac{{2Ey}}{x}$.
The correct answer for this problem is option (A).
Note: If a compound is electrically neutral it means that the elements combining to make the compound have equal and opposite charges. Equivalent mass is defined as the ratio of mass of the metal and mass of the oxygen in the oxide multiplied by 8. It is the mass of a given element which can replace a fixed quantity with another substance.
Complete step by step solution:
It is given in the problem that the metal M and oxygen combine to form a molecular formula ${M_x}{O_y}$ and we need to find the atomic mass of metal M.
Let the atomic mass of the metal M be A, the mass of the metal M in grams is equal to $xA$ grams and the mass of oxygen is equal to $\dfrac{y}{2} \times 32$grams.
The reaction of the given problem is equal,
$ \Rightarrow \underbrace {xM}_{\left( {xA} \right)g} + \underbrace {\left( {\dfrac{y}{2}} \right){O_2}}_{\left( {\dfrac{y}{2} \times 32} \right)} \to {M_x}{O_y}$
The equivalent mass of oxygen is equal to mass of metal M is,
$ \Rightarrow \left( {\dfrac{y}{2} \times 32} \right)g{\text{ of }}{{\text{O}}_2} \to x \times \left( A \right)g{\text{ of M}}$
For 8g ${{\text{O}}_2}$ the equivalent mass of the metal M is equal to,
$ \Rightarrow E = 8g{\text{ of }}{{\text{O}}_2} \to \left( {\dfrac{{x \times A}}{{2y}}} \right)g{\text{ of M}}$
Here E is the equivalent mass.
$ \Rightarrow E = \left( {\dfrac{{x \times A}}{{2y}}} \right)g$
The atomic mass of the metal M is equal to,
$ \Rightarrow A = \dfrac{{2Ey}}{x}$.
The atomic mass of the metal M is equal to $\dfrac{{2Ey}}{x}$.
The correct answer for this problem is option (A).
Note: If a compound is electrically neutral it means that the elements combining to make the compound have equal and opposite charges. Equivalent mass is defined as the ratio of mass of the metal and mass of the oxygen in the oxide multiplied by 8. It is the mass of a given element which can replace a fixed quantity with another substance.
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