Question:
Molar conductivity and conductance of an electrolytic solution are 123.5 Scm$^2$/mole and 0.19 S respectively. Concentration of solution is $x$% w/w, then find the value of x.
(cell constant = 1.3 S cm$^{-1}$; density of solution = 1 gm/ml; molar mass of electrolyte = 75 g/mole)
Molar conductivity and conductance of an electrolytic solution are 123.5 Scm$^2$/mole and 0.19 S respectively. Concentration of solution is $x$% w/w, then find the value of x.
(cell constant = 1.3 S cm$^{-1}$; density of solution = 1 gm/ml; molar mass of electrolyte = 75 g/mole)
(cell constant = 1.3 S cm$^{-1}$; density of solution = 1 gm/ml; molar mass of electrolyte = 75 g/mole)
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Remember: To solve for concentration in solutions, use relations involving conductivity, molar conductivity, and cell constant to derive the mass fraction.
Updated On: Apr 7, 2026
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Correct Answer: 15
Solution and Explanation
Step 1: Write the relation for conductance.
The conductance (\(G\)) of an electrolytic solution is given by the relation: \[ G = \kappa \times \frac{L}{A} \] Where: \(\kappa\) is the molar conductivity, \(L\) is the length, and \(A\) is the area of the electrode.
Step 2: Define the relationship for molar conductivity.
Molar conductivity (\(\Lambda_m\)) is defined as the conductance of the solution for a given concentration: \[ \Lambda_m = \frac{G \cdot V}{n} \] Where \(V\) is the volume of the solution and \(n\) is the number of moles of the solute.
Step 3: Relate given data.
We are given: \(\Lambda_m = 123.5 \, \text{S cm}^2/\text{mole}\), \(G = 0.19 \, \text{S}\), cell constant = \(1.3 \, \text{S cm}^{-1}\), density = \(1 \, \text{gm/ml}\), molar mass = \(75 \, \text{g/mole}\).
Step 4: Use the formula for concentration.
We use the formula for mass fraction concentration: \[ \frac{x}{100} = \frac{m}{\rho V} \] Where \(m\) is the mass of solute, \(\rho\) is the density, and \(V\) is the volume of the solution.
Step 5: Calculate the value of x.
By substituting the known values and solving, we get: \[ x = 15 \, \text{percent} \]
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