A solution is prepared by adding 1.70 g of solid NaCl to 50.0 mL of 0.100 M CaCl₂.What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL. A) 0.782 B) 0.682 C) 0.134 D) 0.234 E) 0.582

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First, calculate the moles of NaCl added: $ \frac{1.70 \, \text{g}}{58.44 \, \text{g/mol}} = 0.0291 \, \text{mol} $. Each mole of NaCl gives 1 mole of Cl⁻. Then, calculate the moles of Cl⁻ from CaCl2: $ 50.0 \, \text{mL} \times 0.100 \, \text{M} = 0.0050 \, \text{mol} $ of CaCl2, which gives $ 0.0050 \, \text{mol} \times 2 = 0.0100 \, \text{mol} $ of Cl⁻. Total moles of Cl⁻ = $ 0.0291 + 0.0100 = 0.0391 \, \text{mol} $. Molarity of Cl⁻ = $ \frac{0.0391 \, \text{mol}}{0.0500 \, \text{L}} = 0.782 \, \text{M} $.

A Solution Is Prepared by Adding 1