haddarh6
contestada

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?

Respuesta :

gerishonwambugu
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

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