Question

The resistance of a 0.5M solution of an electrolyte enclosed between two platinum electrodes 1.5cm apart having an area of 2.0 cm² was found to be 30 ohm. calculate molar conductivity of the solution​

Answer 1

• Explanation:
• The resistance of a 0.5M solution of an electrolyte enclosed between two platinum electrodes 1.5cm apart having an area of 2.0 cm² was found to be 30 ohm. ... 0.5 ohm-1 cm2 mole-1 is the molar conductivity of the solution.

Answer 2

Given:

The resistance of a 0.5 M solution of an electrolyte enclosed between two Platinum electrodes 1.5 cm apart and having an area of cross-section 2.0 cm² was found to be 30 ohms.

To Find:

Calculate the molar conductivity of the solution.

Solution:

We all know that,

Specific conductance={(resistance) * (cross-sectional area)}$^{-1}$

or, к$=\frac{1}{R} *\frac{1}{a}$

or, к=$\frac{1}{30} *\frac{1.5}{2.0}$

Therefore, к=$0.025$Ω$^{-1}$$m^{-1}$

Now, Molar conductance = (1000*к)*$\frac{1}{Concentration}$

or, ∧ₙ $=\frac{1000*0.025}{0.5}$

or, ∧ₙ $=50$Ω$^ { - 1}$m²$mol^{- 1}$

Therefore, the molar conductivity of the solution is 50Ω$^ { - 1}$m²$mol^{- 1}$.