
What is the equivalent mass of ${\text{Mn}}{{\text{O}}_{\text{2}}}$ in the present titration?
a. $\dfrac{{{\text{M}}{\text{W}}}}{{\text{1}}}$
b. $\dfrac{{{\text{M}}{\text{W}}}}{2}$
c. $\dfrac{{{\text{M}}{\text{W}}}}{3}$
d. $\dfrac{2}{3}{\text{M}}{\text{W}}$
Answer
243.6k+ views
Hint: As we know in a chemical reaction, equivalence of a substance is explained as the amount of which combines with 1 mole of hydrogen atoms or replaces the same number of hydrogen atoms. Thus equivalent weight in grams is weight in grams of 1 equivalent.
Complete step by step answer:
We know that given titration takes place in an acidic medium. Thus we can write the reaction of ${\text{Mn}}{{\text{O}}_{\text{2}}}$ is as follows:
${\text{M}}{{\text{n}}^{{\text{4 + }}}} + 2{e^ - } \to {\text{M}}{{\text{n}}^{2 + }}$
In the above reaction, oxidation number of Mn decreases from 4 to 2 [4-2=2]. Therefore the change in oxidation number of Mn is 2.
Now we calculate equivalent weight is as follows:
$
{\text{Equivalent}}\,{\text{weght = }}\dfrac{{{\text{Molecular}}\,{\text{weight}}}}{{{\text{Change}}\,{\text{in}}\,{\text{oxidation}}\,{\text{number}}}} \\
= \dfrac{{{\text{M}}{\text{W}}}}{{\text{2}}} \\
$
Hence option B is the correct answer.
Additional Information: We know that the law of equivalence is one equivalence of an element combined with one equivalent of others. Equivalent weight of an acid in an acid base neutralization reaction is the portion of weight of 1 mole of the acid that can furnish 1 mole of hydrogen ion and equivalent weight of a base is part of weight of one mole of base that can furnish 1 mole of hydroxide ion or accept 1 mole of hydrogen ion.
Note:
As we know for an acid titration is a procedure of quantitative analysis to determine the concentration of an acid by accurately neutralizing it with a standard solution of a base or acid that has known concentration. Generally we use titration in the food industry.
Complete step by step answer:
We know that given titration takes place in an acidic medium. Thus we can write the reaction of ${\text{Mn}}{{\text{O}}_{\text{2}}}$ is as follows:
${\text{M}}{{\text{n}}^{{\text{4 + }}}} + 2{e^ - } \to {\text{M}}{{\text{n}}^{2 + }}$
In the above reaction, oxidation number of Mn decreases from 4 to 2 [4-2=2]. Therefore the change in oxidation number of Mn is 2.
Now we calculate equivalent weight is as follows:
$
{\text{Equivalent}}\,{\text{weght = }}\dfrac{{{\text{Molecular}}\,{\text{weight}}}}{{{\text{Change}}\,{\text{in}}\,{\text{oxidation}}\,{\text{number}}}} \\
= \dfrac{{{\text{M}}{\text{W}}}}{{\text{2}}} \\
$
Hence option B is the correct answer.
Additional Information: We know that the law of equivalence is one equivalence of an element combined with one equivalent of others. Equivalent weight of an acid in an acid base neutralization reaction is the portion of weight of 1 mole of the acid that can furnish 1 mole of hydrogen ion and equivalent weight of a base is part of weight of one mole of base that can furnish 1 mole of hydroxide ion or accept 1 mole of hydrogen ion.
Note:
As we know for an acid titration is a procedure of quantitative analysis to determine the concentration of an acid by accurately neutralizing it with a standard solution of a base or acid that has known concentration. Generally we use titration in the food industry.
Recently Updated Pages
JEE Main 2026 Session 2 City Intimation Slip & Exam Date: Expected Date, Download Link

JEE Main 2026 Session 2 Application Form: Reopened Registration, Dates & Fees

JEE Main 2026 Session 2 Registration (Reopened): Last Date, Fees, Link & Process

WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Know The Difference Between Fluid And Liquid

Difference Between Crystalline and Amorphous Solid: Table & Examples

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Angle of Deviation in a Prism

Understanding Differential Equations: A Complete Guide

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Chemistry Chapter 9 - Hydrocarbons - 2025-26

CBSE Notes Class 11 Chemistry Chapter 5 - Thermodynamics - 2025-26

CBSE Notes Class 11 Chemistry Chapter 6 - Equilibrium - 2025-26

CBSE Notes Class 11 Chemistry Chapter 8 - Organic Chemistry Some Basic Principles And Techniques - 2025-26

