(4-490) Calculate the quantity of electricity in coulombs which is needed to decrease the concentration of 1 L of 1 molar of copper II chloride to 0.25 molar \( (\mathrm{Cu}=63.5) \) a) 144750 C b) 48250 C c) 24125 C d) 72375 C

To decrease the concentration of copper(II) chloride from 1 M to 0.25 M in 1 L of solution, we first calculate the amount of substance to be removed. 1. The initial number of moles of CuCl₂ in 1 L of 1 M solution is 1 mol. 2. The final number of moles of CuCl₂ desired in 0.25 M is 0.25 mol. 3. The amount of CuCl₂ to be removed is 1 mol - 0.25 mol = 0.75 mol. Next, we find the number of moles of Cu²⁺ removed, since the dissociation of CuCl₂ produces one Cu²⁺ ion per formula unit: - Moles of Cu²⁺ removed = 0.75 mol. To convert moles to charge in coulombs, we use Faraday's constant (approximately 96485 C/mol), which represents the charge of one mole of electrons. Each Cu²⁺ contributes two electrons: - Charge (Q) = moles of Cu²⁺ × number of electrons transferred × Faraday's constant = 0.75 mol × 2 × 96485 C/mol = 144750 C. Thus, the quantity of electricity needed is **144750 C** (option a).