Volume (in L) of water in the pool is — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Find the conductivity of pool water after NaCl addition.** After adding NaCl (resistance of solution with NaCl in cell = 8000 Ω): $$\kappa_{\text{total}} = \frac{G^*}{R} = \frac{0.4}{8000} = 5 \times 10^{-5}\ \text{S cm}^{-1}$$ **Step 2: Find conductivity due to NaCl alone.** $$\kappa_{\text{NaCl}} = \kappa_{\text{total}} - \kappa_{\text{water}} = 5 \times 10^{-5} - 4 \times 10^{-5} = 1 \times 10^{-5}\ \text{S cm}^{-1}$$ **Step 3: Find concentration of NaCl in pool.** $\lambda^\circ_{\text{NaCl}} = 125\ \text{S cm}^2\text{mol}^{-1}$ $$C_{\text{NaCl}} = \frac{\kappa_{\text{NaCl}} \times 1000}{\lambda^\circ} = \frac{1 \times 10^{-5} \times 1000}{125} = 8 \times 10^{-5}\ \text{mol/L}$$ **Step 4: Find volume using moles of NaCl added.** Moles of NaCl added: $$n_{\text{NaCl}} = \frac{585\ \text{g}}{58.5\ \text{g/mol}} = 10\ \text{mol}$$ $$V_{\text{pool}} = \frac{n_{\text{NaCl}}}{C_{\text{NaCl}}} = \frac{10}{8 \times 10^{-5}} = 1{,}25{,}000\ \text{L}$$ $$\boxed{\text{Answer: A — Volume of pool} = 1{,}25{,}000\ \text{L}}$$