What is the molality of each of the following solutions?

Question

1)Dissolve 0.610 mol of citric acid, C6H8O7, in 1.35 kg of water.
2) Dissolve 0.200 mg of KBr in 5.50 mL of water.
3) Dissolve 4.05 g of aspirin, C9H8O4, in 145 g of dichloromethane, CH2Cl2.

For some reason I keep forgetting how to find molarity? I know it has to do with converting grams to mols but I'm not sure. If someone could explain it simply and give an example?

Answer 1

1.
m = mols/kg solvent
m = 0.610/1.35 = ?

2.
0.200 mg KBr = 0.0002 g.
mols = grams/molar mass = 0.002/119 = 1.68E-6
m = mols/kg solvent = 1.68E-6/0.0055 = ?

etc.