Question

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?

Answer 1

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!