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MHT CET 202621 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
21 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

MHT CET 2026 · 21 April · Evening Shift · Chemistry · Electrochemistry

The molar conductivity of 0.8\ MBaCl_2 solution at 25^ ^ C is 220\...

The molar conductivity of 0.8 MBaCl20.8\ \text{M}\text{BaCl}_2 solution at 25^{}^{\circ}\text{C} is 220 Ω−1cm2mol−1220\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}. Calculate its conductivity.

Options

  1. A0.176 Ω−1cm−10.176\ \Omega^{-1}\text{cm}^{-1}
  2. B0.226 Ω−1cm−10.226\ \Omega^{-1}\text{cm}^{-1}
  3. C0.252 Ω−1cm−10.252\ \Omega^{-1}\text{cm}^{-1}
  4. D0.152 Ω−1cm−10.152\ \Omega^{-1}\text{cm}^{-1}

Correct answer

A. 0.176 Ω−1cm−10.176\ \Omega^{-1}\text{cm}^{-1}

Step-by-step solution

Molar conductivity is given by the formula:

Λm=κ×1000M\Lambda_m = \dfrac{\kappa \times 1000}{M}

where κ\kappa is the conductivity and MM is the molarity of the solution.

Rearranging the formula to solve for κ\kappa:

κ=Λm×M1000\kappa = \dfrac{\Lambda_m \times M}{1000}

Substituting the given values Λm=220 Ω−1cm2mol−1\Lambda_m = 220\ \Omega^{-1}\text{cm}^2\text{mol}^{-1} and M=0.8 MM = 0.8\ \text{M}:

κ=220×0.81000\kappa = \dfrac{220 \times 0.8}{1000}

κ=1761000=0.176 Ω−1cm−1\kappa = \dfrac{176}{1000} = 0.176\ \Omega^{-1}\text{cm}^{-1}

Answer: 0.176 Ω−1cm−10.176\ \Omega^{-1}\text{cm}^{-1}
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