Calculate the conductivity of 0.025 M BaCl_2 solution if the molar conductivity of salt at 25^ ^ C is 325\ ^ -1 \ cm^2\ mol^ -1 .

Correct answer: B. 8.125 × 10^ -3 \ ^ -1 cm^ -1 Given: Concentration, C = 0.025\ M Molar conductivity, _m = 325\ ^ -1 \ cm^2\ mol^ -1 The relationship between molar conductivity and conductivity ( ) is given by: _m = ( × 1000)/(C) Rearranging the formula to solve for : = ( _m × C)/(1000) Substituting the given values: = (325 × 0.025)/(1000) = (8.125)/(1000) = 8.125 × 10^ -3 \ ^ -1 \ cm^ -1 Answer: 8.125 × 10^ -3 \ ^ -1 cm^ -1

MHT CET 202626 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 26 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Electrochemistry

MHT CET 2026 · 26 April · Morning Shift · Chemistry · Electrochemistry

Calculate the conductivity of 0.025 M BaCl_2 solution if the molar...

Calculate the conductivity of

0.025

\text{BaCl}_2

solution if the molar conductivity of salt at

25^{}^{\circ}\text{C}

325\ \Omega^{-1}\ \text{cm}^2\ \text{mol}^{-1}

Options

A $8.125 \times 10^{-2}\ \Omega^{-1}\text{cm}^{-1}$
B $8.125 \times 10^{-3}\ \Omega^{-1}\text{cm}^{-1}$
C $7.6923 \times 10^{-2}\ \Omega^{-1}\text{cm}^{-1}$
D $8.372 \times 10^{-2}\ \Omega^{-1}\text{cm}^{-1}$

Correct answer

B. $8.125 \times 10^{-3}\ \Omega^{-1}\text{cm}^{-1}$

Step-by-step solution

Given:
Concentration,

C = 0.025\ \text{M}

Molar conductivity,

\Lambda_m = 325\ \Omega^{-1}\ \text{cm}^2\ \text{mol}^{-1}

The relationship between molar conductivity and conductivity (\kappa

) is given by:<br>\Lambda_m = \dfrac{\kappa \times 1000}{C}

Rearranging the formula to solve for

\kappa

\kappa = \dfrac{\Lambda_m \times C}{1000}

Substituting the given values:

\kappa = \dfrac{325 \times 0.025}{1000}

\kappa = \dfrac{8.125}{1000}

\kappa = 8.125 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}

Answer:

8.125 \times 10^{-3}\ \Omega^{-1}\text{cm}^{-1}

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