
A 5.0 m mol d \[{m^{ - 3}}\] aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 \[c{m^{ - 1}}\]. What is the molar conductivity of this solution ?(Round off to the nearest integer).
Answer
219.9k+ views
Hint: Molar conductivity is the conductance property of a solution containing one mole of an electrolyte. It is a function of the ionic strength of a solution or the concentration of the particular ionic salt. In other words, molar conductivity can be defined as the conducting power of all the ions that are formed by dissolving a unit mole of an electrolyte in a solution. Molar conductivity is a property of an electrolyte solution that is used to determine the efficiency with which an electrolyte conducts electricity in a solution. Hence, it is not a constant.
Complete Step by Step Solution:
\[{\lambda _M} = \dfrac{K}{C}\] is the equation used to denote the molar conductivity of a solution. Here, “K” stand for the specific conductivity and “c” stands for the concentration in mole per litre.
\[{G_{KCl}}\] = 0.55 mS = 55 \[ \times {10^{ - 5}}\]s
Cell constant = \[\dfrac{I}{A} = 1.3c{m^{ - 1}}\]
\[{\lambda _M}\] = ??
\[K = G(\dfrac{I}{A}) = 55 \times {10^{ - 5 \times 1.3}}S\]\[c{m^{ - 1}}\]
\[{\lambda _M} = \dfrac{{K \times 1000}}{{Molarity}} = \dfrac{{55 \times 1.3 \times {{10}^{ - 5}} \times 1000}}{{5 \times {{10}^{ - 3}}}}\]
\[{\lambda _M} = 11 \times 1.3 \times 10 = 11 \times 13 = 143\]S \[c{m^2}\] \[mo{l^{ - 1}}\]
\[{\lambda _M} = \dfrac{{143 \times 1000 \times {{10}^{ - 3}}S}}{{{{({{10}^{ - 2}}m)}^{ - 2}}}}mo{l^{ - 1}}\]
\[{\lambda _M} = 143 \times 1000 \times {10^{ - 4}}(ms){m^2}mo{l^{ - 1}}\]
\[{\lambda _M} = 14.3(ms){m^2}mo{l^{ - 1}}\]
Therefore, after rounding off to the nearest integer we get the answer as \[{\lambda _M} = 14(ms){m^2}mo{l^{ - 1}}\].
Note: The conducting power of all the ions created by dissolving one mole of an electrolyte in the solution is the general definition of molar conductivity. With a fall in concentration or an increase in dilution, the molar conductivity of both weak and strong electrolytes increases. We know that molar conductivity is the conductivity offered by a unit mole of ions. Even after dilution, considering the same unit mole of ions are present in the solution, the increased dilution results in the dissociation of more electrolytes into ions, and effectively the number of active ions in the solution is increased. Therefore, these active ions impart more conductivity.
Complete Step by Step Solution:
\[{\lambda _M} = \dfrac{K}{C}\] is the equation used to denote the molar conductivity of a solution. Here, “K” stand for the specific conductivity and “c” stands for the concentration in mole per litre.
\[{G_{KCl}}\] = 0.55 mS = 55 \[ \times {10^{ - 5}}\]s
Cell constant = \[\dfrac{I}{A} = 1.3c{m^{ - 1}}\]
\[{\lambda _M}\] = ??
\[K = G(\dfrac{I}{A}) = 55 \times {10^{ - 5 \times 1.3}}S\]\[c{m^{ - 1}}\]
\[{\lambda _M} = \dfrac{{K \times 1000}}{{Molarity}} = \dfrac{{55 \times 1.3 \times {{10}^{ - 5}} \times 1000}}{{5 \times {{10}^{ - 3}}}}\]
\[{\lambda _M} = 11 \times 1.3 \times 10 = 11 \times 13 = 143\]S \[c{m^2}\] \[mo{l^{ - 1}}\]
\[{\lambda _M} = \dfrac{{143 \times 1000 \times {{10}^{ - 3}}S}}{{{{({{10}^{ - 2}}m)}^{ - 2}}}}mo{l^{ - 1}}\]
\[{\lambda _M} = 143 \times 1000 \times {10^{ - 4}}(ms){m^2}mo{l^{ - 1}}\]
\[{\lambda _M} = 14.3(ms){m^2}mo{l^{ - 1}}\]
Therefore, after rounding off to the nearest integer we get the answer as \[{\lambda _M} = 14(ms){m^2}mo{l^{ - 1}}\].
Note: The conducting power of all the ions created by dissolving one mole of an electrolyte in the solution is the general definition of molar conductivity. With a fall in concentration or an increase in dilution, the molar conductivity of both weak and strong electrolytes increases. We know that molar conductivity is the conductivity offered by a unit mole of ions. Even after dilution, considering the same unit mole of ions are present in the solution, the increased dilution results in the dissociation of more electrolytes into ions, and effectively the number of active ions in the solution is increased. Therefore, these active ions impart more conductivity.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
Solutions Class 12 Chemistry Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26

The D and F Block Elements Class 12 Chemistry Chapter 4 CBSE Notes - 2025-26

NCERT Solutions for Class 12 Chemistry Chapter Chapter 7 Alcohol Phenol and Ether

NCERT Solutions ForClass 12 Chemistry Chapter Chapter 8 Aldehydes Ketones And Carboxylic Acids

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

