Back to Directory
NEET CHEMISTRYEasy

The conductivity of centimolar solution of KCl at 25∘C25^{\circ}\text{C} is 0.0210 ohm–1 cm–10.0210\text{ ohm}^{–1}\text{ cm}^{–1} and the resistance of the cell containing the solution at 25∘C25^{\circ}\text{C} is 60 ohm60\text{ ohm}. The value of the cell constant is:

A

3.34 cm–13.34\text{ cm}^{–1}

B

1.34 cm–11.34\text{ cm}^{–1}

C

3.28 cm–13.28\text{ cm}^{–1}

D

1.26 cm–11.26\text{ cm}^{–1}

Step-by-Step Solution

Given: Conductivity (κ\kappa) = 0.0210 \Omega−1 cm−10.0210\text{ \Omega}^{-1}\text{ cm}^{-1} Resistance (RR) = 60 \Omega60\text{ \Omega}

The relation between cell constant (G∗G^*), conductivity (κ\kappa), and resistance (RR) is: Cell constant (G∗)=Conductivity (κ)×Resistance (R)\text{Cell constant } (G^*) = \text{Conductivity } (\kappa) \times \text{Resistance } (R)

Substituting the given values into the formula: G∗=0.0210 \Omega−1 cm−1×60 \OmegaG^* = 0.0210\text{ \Omega}^{-1}\text{ cm}^{-1} \times 60\text{ \Omega} G∗=1.26 cm−1G^* = 1.26\text{ cm}^{-1}

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started