Question

The molar conductivity of \(0.5 \, \text{mol/dm}^3\) solution of \(AgNO_3\) with electrolytic conductivity of \(5.76 \times 10^{-3} \, \text{S cm}^{-1}\) at \(298\,K\) is:

• A. \(0.086 \, \text{S cm}^2\text{/mol}\)
• B. \(28.8 \, \text{S cm}^2\text{/mol}\)
• C. \(2.88 \, \text{S cm}^2\text{/mol}\)
• D. \(11.52 \, \text{S cm}^2\text{/mol}\)

Hint:

While calculating molar conductivity, always remember the formula: \[ \Lambda_m = \frac{\kappa \times 1000}{C} \] where concentration must be in \(\text{mol L}^{-1}\).

Answer 1

The Correct Option is D

Concept: Molar conductivity (\(\Lambda_m\)) is defined as the conductance of all ions produced by one mole of an electrolyte dissolved in a given volume of solution. The relation between molar conductivity and conductivity is: \[ \Lambda_m = \frac{\kappa \times 1000}{C} \] where:
• \(\Lambda_m\) = molar conductivity \((\text{S cm}^2\text{/mol})\)
• \(\kappa\) = conductivity \((\text{S cm}^{-1})\)
• \(C\) = concentration in \((\text{mol L}^{-1})\)

Step 1: Write the given data Given: \[ \kappa = 5.76 \times 10^{-3} \, \text{S cm}^{-1} \] \[ C = 0.5 \, \text{mol dm}^{-3} \] Since: \[ 1 \, \text{dm}^3 = 1 \, \text{L} \] therefore: \[ C = 0.5 \, \text{mol L}^{-1} \]

Step 2: Apply the molar conductivity formula \[ \Lambda_m = \frac{\kappa \times 1000}{C} \] Substitute values: \[ \Lambda_m = \frac{5.76 \times 10^{-3} \times 1000}{0.5} \]

Step 3: Simplify the numerator \[ 5.76 \times 10^{-3} \times 1000 \] Since: \[ 1000 = 10^3 \] \[ 5.76 \times 10^{-3} \times 10^3 = 5.76 \] Thus: \[ \Lambda_m = \frac{5.76}{0.5} \]

Step 4: Perform the division \[ \frac{5.76}{0.5} = 11.52 \] Therefore: \[ \Lambda_m = 11.52 \, \text{S cm}^2\text{/mol} \] Final Answer: \[ \boxed{11.52 \, \text{S cm}^2\text{/mol}} \]