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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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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?

User Aric
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1 Answer

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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
User Gagarwa
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