Answer
Verified
112.8k+ views
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key
JEE Main 2023 (January 30th Shift 1) Physics Question Paper with Answer Key
JEE Main 2023 (January 25th Shift 1) Physics Question Paper with Answer Key
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Angle of Deviation in Prism - Important Formula with Solved Problems for JEE
Average and RMS Value for JEE Main
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Inductive Effect and Acidic Strength - Types, Relation and Applications for JEE
Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction
NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry
NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium
Equilibrium Class 11 Notes: CBSE Chemistry Chapter 6
Thermodynamics Class 11 Notes: CBSE Chapter 5