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500 ml of 0.1 M AlCl(3) is mixed with 50...

500 ml of `0.1 M AlCl_(3)` is mixed with 500 ml of `0.1 M MgCl_(2)` solution. Then calculate the molarity of `Cl^(-)` in final solution.

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To solve the problem, we need to calculate the total moles of Cl⁻ ions contributed by both AlCl₃ and MgCl₂ solutions and then find the molarity of Cl⁻ in the final mixed solution. ### Step-by-Step Solution: 1. **Calculate the moles of Cl⁻ from AlCl₃:** - The formula for aluminum chloride is AlCl₃, which dissociates in solution to give 3 Cl⁻ ions. - First, we calculate the moles of AlCl₃ in the 500 ml of 0.1 M solution: \[ ...
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