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2. How many grams of potassium iodide are needed to make 125 mL of a 1.28 M KI solution? (1.5)

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Answer #1

125 mL of a 128 M KI 1000 ml contains. 1.28 mole ki 125 mL contains 1.28x 125 - - 0.16 mole kI - 1000 Mass of kI needed = 0.1

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