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Step-by-Step Solution
Step 1: Identify the known quantities
• Cell constant (denoted as K_\text{cell} ): 1.14 cm−1
• Resistance ( R ) of the cell: 1500 Ω
• Concentration of KCl solution ( c ): 0.001 M
Step 2: Write the formula for conductivity
Conductivity ( \kappa ) of the solution is related to the resistance and the cell constant as:
\kappa = \dfrac{K_\text{cell}}{R}
Here, K_\text{cell} = \dfrac{l}{A} is given directly as 1.14 cm−1, and R = 1500 Ω.
Step 3: Calculate the conductivity
Substitute the values into the formula:
\kappa = \dfrac{1.14}{1500} \,\text{S cm}^{-1}
Compute this value:
\kappa = 7.6 \times 10^{-4} \,\text{S cm}^{-1}
Step 4: Write the formula for molar conductivity
Molar conductivity ( \Lambda_m ) is given by:
\Lambda_m = \dfrac{1000 \times \kappa}{c}
Here, c is in mol/cm3. Often, we convert c from M (which is mol/L) to mol/cm3.
• 1 M = 1 mol/L = 1 mol/(1000 mL) = 1 mol/(1000 cm3)
• Hence, 0.001 M = 0.001 mol/(1000 cm3) = 1 \times 10^{-6} mol/cm3.
In practice, we commonly use the form:
\Lambda_m (\text{in S cm}^2 \text{ mol}^{-1}) = 1000 \times \dfrac{\kappa (\text{in S cm}^{-1})}{c (\text{in mol L}^{-1})}
because multiplying by 1000 accounts for the conversion from L to mL (or equivalently to cm3) when c is in mol/L.
Step 5: Substitute the values into the molar conductivity formula
Now, we have:
• \kappa = \dfrac{1.14}{1500} \,\text{S cm}^{-1}
• c = 0.001 \,\text{mol L}^{-1}
Therefore,
\Lambda_m = 1000 \times \dfrac{\left(\dfrac{1.14}{1500}\right)}{0.001} \,\text{S cm}^2 \text{mol}^{-1}
Simplify step by step:
1) Inside the numerator: \dfrac{1.14}{1500} = 7.6 \times 10^{-4} (approximately)
2) Divide by 0.001: \dfrac{7.6 \times 10^{-4}}{0.001} = 0.76
3) Multiply by 1000: 1000 \times 0.76 = 760
Hence, the molar conductivity is:
\Lambda_m = 760 \,\text{S cm}^2 \,\text{mol}^{-1}
Final Answer
760
