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MHT CET 202623 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
23 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

MHT CET 2026 · 23 April · Morning Shift · Chemistry · Electrochemistry

Find the molar conductivity of 0.001 M AgNO_3 solution if the...

Find the molar conductivity of 0.0010.001M AgNO3\text{AgNO}_3 solution if the conductivity of it is 1.2×10−4Ω−11.2 \times 10^{-4}\Omega^{-1} cm^{-1}..

Options

  1. A120Ω−1120\Omega^{-1} cm^2mol−1mol^{-1}.
  2. B130Ω−1130\Omega^{-1} cm^2mol−1mol^{-1}.
  3. C140Ω−1140\Omega^{-1} cm^2mol−1mol^{-1}.
  4. D150Ω−1150\Omega^{-1} cm^2mol−1mol^{-1}.

Correct answer

A. 120Ω−1120\Omega^{-1} cm^2mol−1mol^{-1}.

Step-by-step solution

Given:

Concentration, M=0.001 MM = 0.001 \text{ M}

Conductivity, κ=1.2×10−4 Ω−1 cm−1\kappa = 1.2 \times 10^{-4} \ \Omega^{-1} \text{ cm}^{-1}

The formula for molar conductivity Λm\Lambda_m is:

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

Substituting the given values:

Λm=1.2×10−4×10000.001\Lambda_m = \dfrac{1.2 \times 10^{-4} \times 1000}{0.001}

Λm=1.2×10−110−3\Lambda_m = \dfrac{1.2 \times 10^{-1}}{10^{-3}}

Λm=1.2×102\Lambda_m = 1.2 \times 10^2

Λm=120 Ω−1 cm2 mol−1\Lambda_m = 120 \ \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}
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