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Step-by-Step Solution
Step 1: Identify the Given Data
• Molar concentration of KCl,
C = 5.0 \text{ mmol dm}^{-3} = 5.0 \times 10^{-3} \text{ mol dm}^{-3}.
• Measured conductance of the solution, G = 0.55 \text{ mS} = 0.55 \times 10^{-3} \text{ S}.
• Cell constant, k = 1.3 \text{ cm}^{-1}.
Step 2: Convert Conductance to Conductivity
Conductivity ( \kappa ) is given by:
\kappa = G \times \text{(Cell constant)}.
Substituting the values:
\kappa = (0.55 \times 10^{-3} \,\text{S}) \times (1.3 \,\text{cm}^{-1})
= 0.715 \times 10^{-3} \,\text{S cm}^{-1}
= 7.15 \times 10^{-4} \,\text{S cm}^{-1}.
Step 3: Use the Formula for Molar Conductivity
Molar conductivity ( \Lambda_m ) can be calculated using the relation:
\Lambda_m \,(\text{in } \text{S cm}^2 \text{ mol}^{-1})
= \frac{\kappa \times 1000}{C},
where C is the concentration in \text{mol dm}^{-3} , and \kappa is in \text{S cm}^{-1} .
Step 4: Substitute Numerical Values
\Lambda_m = \frac{(7.15 \times 10^{-4} \,\text{S cm}^{-1}) \times 1000}{5.0 \times 10^{-3} \,\text{mol dm}^{-3}}.
Simplify:
\Lambda_m = \frac{7.15 \times 10^{-1} \,\text{S cm}^{-1}}{5.0 \times 10^{-3} \,\text{mol dm}^{-3}}
= \frac{0.715 \,\text{S cm}^{-1}}{5.0 \times 10^{-3}}
= 143 \,\text{S cm}^2 \text{mol}^{-1}.
Step 5: Convert to the Desired Units (mS m2 mol−1)
1 S cm2 mol−1 = 0.1 mS m2 mol−1 (because 1 cm = 0.01 m and 1 S = 1000 mS).
Thus,
143 \,\text{S cm}^2 \text{mol}^{-1}
= 14.3 \,\text{mS m}^2 \text{mol}^{-1}.
Rounded off to the nearest integer gives
14 \,\text{mS m}^2 \text{mol}^{-1}.
Final Answer
The molar conductivity of the solution is 14 mS m2 mol−1.
