Find the cell constant for a conductance cell in which the conductance, G, of a 0.100 \(\mathrm{M} \mathrm{KCl}\) solution is 0.01178 at \(25^{\circ} \mathrm{C}\). The molar conductance for \(0.100 \mathrm{M} \mathrm{KCl}\) at \(25^{\circ} \mathrm{C}\) is \(128.96 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\). If a \(0.0500 \mathrm{M}\) solution of an electrolyte has a measured conductance of \(0.00824 \mathrm{~S}\) using this cell, what is the molar conductance of the electrolyte?

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### Calculation of Cell Constant

**Given Data for \(0.100 \, \text{M} \, \text{KCl}\):**

– Conductance, \( G = 0.01178 \, \text{S} \)

– Molar conductance, \( \Lambda_m = 128.96 \, \text{S cm}^2 \text{mol}^{-1} \)

– Concentration, \( C = 0.100 \, \text{M} \)

**Formula for Cell Constant:**

\[ K_{\text{cell}} = \frac{\Lambda_m}{G/C} \]

**Calculation:**

\[ K_{\text{cell}} = \frac{128.96 \, \text{S cm}^2 \text{mol}^{-1}}{0.01178 \, \text{S} / 0.100 \, \text{M}} \]

\[ K_{\text{cell}} = \frac{128.96}{0.1178} \]

\[ K_{\text{cell}} \approx 1094.57 \, \text{cm}^{-1} \]

### Calculation of Molar Conductance of the Electrolyte

**Given Data for Electrolyte Solution:**

– Conductance, \( G = 0.00824 \, \text{S} \)

– Concentration, \( C = 0.0500 \, \text{M} \)

**Formula for Molar Conductance:**

\[ \Lambda_m = \frac{G}{C} \times K_{\text{cell}} \]

**Calculation:**

\[ \Lambda_m = \frac{0.00824 \, \text{S}}{0.0500 \, \text{M}} \times 1094.57 \, \text{cm}^{-1} \]

\[ \Lambda_m = 0.1648 \times 1094.57 \]

\[ \Lambda_m \approx 180.39 \, \text{S cm}^2 \text{mol}^{-1} \]

**Result:**

The molar conductance of the \(0.0500 \, \text{M}\) electrolyte solution is approximately \(180.39 \, \text{S cm}^2 \text{mol}^{-1}\).