Question

The molar conductivity of a 0.5 mol dm⁻³ solution of AgNO₃ with specific conductance 5.76 × 10⁻³ S cm⁻¹ at 298 K is:

• A. 0.086 S cm² mol⁻¹
• B. 2.88 S cm² mol⁻¹
• C. 28.8 S cm² mol⁻¹
• D. 11.52 S cm² mol⁻¹

Hint:

Always check units! If κ is given in S m⁻¹, the formula changes to = (κ)/(1000 × M). Since the options here use cm², the standard formula used in Step 2 is the correct choice.

Answer 1

The Correct Option is D

Concept: Molar conductivity () is the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution. It is related to specific conductance (kappa, κ) and molarity (M) by the following formula: = (κ × 1000)/(M) Where: • κ is the specific conductance in S cm⁻¹. • M is the molarity in mol L⁻¹ (which is equivalent to mol dm⁻³).
• Specific conductance, κ = 5.76 × 10⁻³ S cm⁻¹ • Molarity (Concentration), M = 0.5 mol dm⁻³
Substituting the values into the formula: = 5.76 × 10⁻³ × 10000.5 = (5.76)/(0.5) = 11.52 S cm² mol⁻¹