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)

Hint:

Remember: To solve for concentration in solutions, use relations involving conductivity, molar conductivity, and cell constant to derive the mass fraction.

Answer 1

Correct Answer: 15

Step 1: Understanding the Concept:
Conductivity (\(\kappa\)) is the ability of a solution to conduct electricity and is related to the measured conductance (\(G\)) and the cell constant (\(G^*\)) by: \(\kappa = G \times G^*\).
Molar conductivity (\(\Lambda_m\)) relates conductivity to molar concentration: \(\Lambda_m = \frac{\kappa \times 1000}{M}\) (where M is molarity in mol/L).
Once we determine molarity, it can be converted to weight percentage using density and molar mass.
Step 2: Key Formula or Approach:
1. \(\kappa = G \times G^*\)
2. \(M = \frac{\kappa \times 1000}{\Lambda_m}\)
3. \(% w/w = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\)
Step 3: Detailed Explanation:
Step 3.1: Calculate Conductivity (\(\kappa\))
\[ \kappa = G \times G^* = 0.19 \text{ S} \times 1.3 \text{ cm}^{-1} = 0.247 \text{ S cm}^{-1} \]
Step 3.2: Calculate Molarity (\(M\))
\[ M = \frac{\kappa \times 1000}{\Lambda_m} = \frac{0.247 \times 1000}{123.5} = \frac{247}{123.5} = 2 \text{ mol/L} \]
Step 3.3: Convert Molarity to % w/w
Since density \(d = 1\) g/ml, mass of 1000 ml (= 1 L) of solution = 1000 g.
Number of moles of solute in 1 L = 2 mol.
Mass of solute = \(2 \times 75 = 150\) g.
\[ % w/w = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 = \frac{150}{1000} \times 100 = 15% \]
Step 4: Final Answer:
The value of x is 15.