Molality questions. Can you break it down for me

I am having a little trouble with a Question on this homework and hope you can explain it, here they are:

1.) In order to Prepare a .523 m(olality) aqueous solution of Potassium Iodide (KI), how many grams of potassium Iodide must be added to 2.00 kg of water?

molality = (# of moles of solute)/(weight of solvent in kg)

.523 = x/2
x = 1.046 moles

KI = 166 grams per mole

166/1 = x/1.046
x = 173.64 grams