Question 22 of 29advanced🔧 ApplyNumerical5 marks

The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M: 0.001, 0.010, 0.020, 0.050, 0.100; 10² × κ/S m⁻¹: 1.237, 11.85, 23.15, 55.53, 106.74. Calculate Λm for all concentrations and draw a plot between Λm and c^½. Find the value of Λm⁰.

Correct Answer

Λm values (S cm² mol⁻¹): 123.7, 118.5, 115.75, 111.06, 106.74 for concentrations 0.001, 0.010, 0.020, 0.050, 0.100 M respectively. Λm⁰ ≈\approx 124.0 S cm² mol⁻¹

Exercise: EXERCISES | Q: 2.10 | (Page Page 30)
For More Understanding

Explanation

This question tests the calculation of molar conductivity from conductivity and the Debye-Hückel-Onsager equation. The plot of Λm vs c^½ is a straight line for strong electrolytes, and extrapolation to c^½ = 0 gives the limiting molar conductivity Λm⁰.

Solution Steps

  1. Step 1: Convert κ values - Given κ = (value) × 10⁻² S m⁻¹. For 0.001 M: κ = 1.237×101.237 \times 10⁻² S m⁻¹; 0.010 M: κ = 11.85×1011.85 \times 10⁻² S m⁻¹; 0.020 M: κ = 23.15×1023.15 \times 10⁻² S m⁻¹; 0.050 M: κ = 55.53×1055.53 \times 10⁻² S m⁻¹; 0.100 M: κ = 106.74×10106.74 \times 10⁻² S m⁻¹

  2. Step 2: Convert concentration - c (mol m⁻³) = c (mol L⁻¹) × 1000. For 0.001 M: c = 1 mol m⁻³; 0.010 M: c = 10 mol m⁻³; 0.020 M: c = 20 mol m⁻³; 0.050 M: c = 50 mol m⁻³; 0.100 M: c = 100 mol m⁻³

  3. Step 3: Calculate Λm - Using Λm = κ/c. For 0.001 M: Λm = (1.237×101.237 \times 10⁻²)/(1) = 1.237×101.237 \times 10⁻² S m² mol⁻¹ = 123.7 S cm² mol⁻¹. Similarly, 0.010 M: Λm = 118.5 S cm² mol⁻¹; 0.020 M: Λm = 115.75 S cm² mol⁻¹; 0.050 M: Λm = 111.06 S cm² mol⁻¹; 0.100 M: Λm = 106.74 S cm² mol⁻¹

  4. Step 4: Calculate c^½ - For 0.001 M: c^½ = 0.0316; 0.010 M: c^½ = 0.100; 0.020 M: c^½ = 0.141; 0.050 M: c^½ = 0.224; 0.100 M: c^½ = 0.316 (all in (mol L⁻¹)^½)

  5. Step 5: Plot and find Λm⁰ - Plot Λm (y-axis) vs c^½ (x-axis). The plot is a straight line. From the intercept at c^½ = 0, Λm⁰ ≈\approx 124.0 S cm² mol⁻¹

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