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Redox Reactions and Electrochemistry: Chemistry | JEE Main

Q: Why is the factor 1000 in the numerator for ?

Because molarity is in moles per liter ( ), so the concentration in is . Since , dividing by puts into the numerator.

The resistivity of a

0.8\mathrm{M}

solution of an electrolyte is

5 \times 10^{-3}\Omega\mathrm{cm}

. Its molar conductivity is

\underline{\quad\quad\quad} \times 10^4\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}

(Nearest integer)

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Step-by-step solutionView

Correct answer

The molar conductivity of the solution is $25 \times 10^4\,\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ , so the integer to enter is 25.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Molarity of the electrolyte solution $M = 0.8\text{ M} = 0.8\text{ mol/L}$ , resistivity $\rho = 5 \times 10^{-3}\,\Omega\text{ cm}$ .

02find

Molar conductivity $\Lambda_m$ in units of $\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ , expressed as $x \times 10^4$ .

03strategise

Conductivity $\kappa$ is the reciprocal of resistivity: $\kappa = \frac{1}{\rho}$ . The molar conductivity is related to conductivity and molarity by $\Lambda_m = \frac{\kappa \times 1000}{M}$ , where $\kappa$ is in $\Omega^{-1}\text{ cm}^{-1}$ and $M$ is in $\text{mol/L}$ .

04execute

Substitute $\kappa = \frac{1}{5 \times 10^{-3}}\,\Omega^{-1}\text{ cm}^{-1}$ and $M = 0.8$ into the formula: $\Lambda_m = \frac{1}{5 \times 10^{-3}} \times \frac{1000}{0.8} = \frac{10^3}{5 \times 10^{-3} \times 0.8} = \frac{10^3}{4 \times 10^{-3}} = \frac{10^6}{4} = 2.5 \times 10^5 = 25 \times 10^4\,\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$

✓verify

Check units: $\kappa = 200\,\Omega^{-1}\text{ cm}^{-1}$ . $\Lambda_m = \frac{200 \times 1000}{0.8} = 250{,}000 = 25 \times 10^4\,\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ . The requested factor is $\times 10^4$ , giving the integer 25.

Hints that build this answer step by step

What is the relationship between conductivity ( $\kappa$ ) and resistivity ( $\rho$ ), and how is molar conductivity ( $\Lambda_m$ ) calculated when concentration $C$ is in $\text{mol L}^{-1}$ and $\kappa$ is in $\Omega^{-1}\text{ cm}^{-1}$ ?
$\kappa = \frac{1}{\rho}$ and $\Lambda_m = \frac{1000 \times \kappa}{C}$
What is the value of conductivity $\kappa$ in $\Omega^{-1}\text{ cm}^{-1}$ given $\rho = 5 \times 10^{-3}\,\Omega\text{ cm}$ ?
$\kappa = 200\,\Omega^{-1}\text{ cm}^{-1}$
Substitute $\kappa = 200\,\Omega^{-1}\text{ cm}^{-1}$ and $C = 0.8\,\text{M}$ into $\Lambda_m = \frac{1000 \times \kappa}{C}$ to find the integer multiplier for $10^4\,\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ .
25

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the factor 1000 in the numerator for

\Lambda_m = \frac{\kappa \times 1000}{M}

Because molarity $M$ is in moles per liter ( $1000\text{ cm}^3$ ), so the concentration in $\text{mol/cm}^3$ is $M / 1000$ . Since $\Lambda_m = \kappa / c$ , dividing by $(M/1000)$ puts $1000$ into the numerator.

Answer

The molar conductivity of the solution is $25 \times 10^4\,\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ , so the integer to enter is 25.

Why each option works or fails

Step-by-step solution

given: Molarity of the electrolyte solution $M = 0.8\text{ M} = 0.8\text{ mol/L}$ , resistivity $\rho = 5 \times 10^{-3}\,\Omega\text{ cm}$ .
find: Molar conductivity $\Lambda_m$ in units of $\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ , expressed as $x \times 10^4$ .
strategise: Conductivity $\kappa$ is the reciprocal of resistivity: $\kappa = \frac{1}{\rho}$ . The molar conductivity is related to conductivity and molarity by $\Lambda_m = \frac{\kappa \times 1000}{M}$ , where $\kappa$ is in $\Omega^{-1}\text{ cm}^{-1}$ and $M$ is in $\text{mol/L}$ .
execute: Substitute $\kappa = \frac{1}{5 \times 10^{-3}}\,\Omega^{-1}\text{ cm}^{-1}$ and $M = 0.8$ into the formula: $\Lambda_m = \frac{1}{5 \times 10^{-3}} \times \frac{1000}{0.8} = \frac{10^3}{5 \times 10^{-3} \times 0.8} = \frac{10^3}{4 \times 10^{-3}} = \frac{10^6}{4} = 2.5 \times 10^5 = 25 \times 10^4\,\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$
verify: Check units: $\kappa = 200\,\Omega^{-1}\text{ cm}^{-1}$ . $\Lambda_m = \frac{200 \times 1000}{0.8} = 250{,}000 = 25 \times 10^4\,\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ . The requested factor is $\times 10^4$ , giving the integer 25.

Redox Reactions and Electrochemistry: Chemistry | JEE Main

What feels right?

The molar conductivity of the solution is $25 \times 10^4\,\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ , so the integer to enter is 25.

Why each option works or fails

Solution

01given

02find

03strategise

04execute

✓verify

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Redox Reactions and Electrochemistry: Chemistry | JEE Main

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