Verified Solution • 2025–26 Session moderate

QJEEM21Cha68e: JEE Main 2021 Chemistry - Class 12 JEEM21Cha68e Solution: If the conductivity of mercury at 0 C is 1.07 10 6 S m 1 and…

If the conductivity of mercury at 0 ∘^\circ C is 1.07 ×\times 106 S m −- 1 and the resistance of a cell containing mercury is 0.243 Ω\Omega , then the cell constant of the cell is x ×\times 104 m −- 1. The value of x is ____________. (Nearest integer)

Step-by-Step Verified Solution

Conductance (G) is reciprocal of resistance (R)

R=1GR = {1 \over G} or G=1R=10.243Ω=4.115Ω−1G = {1 \over R} = {1 \over {0.243\Omega }} = 4.115{\Omega ^{ - 1}}

Relation between conductance (G),

conductivity ( κ\kappa ) and cell constant (lA)\left( {{l \over A}} \right) is given as

κ=GlA\kappa = {{Gl} \over A}

⇒lA=κG=1.07×106Sm−14.115Ω−1=26×104m−1⇒x=26\Rightarrow {l \over A} = {\kappa \over G} = {{1.07 \times {{10}^6}S{m^{ - 1}}} \over {4.115{\Omega ^{ - 1}}}} = 26 \times {10^4}{m^{ - 1}} \Rightarrow x = 26

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