Question

Im having trouble with a chemistry question. The question is as follows

In order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.kg of water?

Am I correct in thinking that the potassium iodide is the solute and do I use mass of solute / volute of solvent?

What is the little m indicating

Answer 1

molality = moles solute / kg solvent

0.523 = moles / 2.0 kg

0.523 x 2.0 => 1.046 moles

Molar mass potassium iodide ( KI ) = 166.0028 g/mol

1 mole KI ------------------ 166.0028 g
1.046 moles -------------- ?? ( mass KI )

1.046 x 166.0028 / 1 => 173.638 g of KI

hope this helps!