In order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.00 kg of water?

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asked Jan 25, 2018 212k views

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Gagarwa · Answer 1 · 2018-01-29T02:31:44+0000

M = amount of the solute / mass of the solvent

0.523 = x / 2.00

x = 0.523 * 2.00

x = 1,046 moles

molar mass KI = 166.0028 g/mole

Mass = 1,046 * 166.0028

Mass ≈ 173.63 g

In order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.00 kg of water?

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