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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 2.00kg of water?
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Jul 8, 2017
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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 2.00kg of water?
Chemistry
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MKMohanty
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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/mol
Mass = 1,046 * 166.0028
Mass
≈
173.63 g
hope this helps!
Lmocsi
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Jul 14, 2017
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Lmocsi
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