How would you prepare 250 ml of a 0.040 M ammonium chloride solution?

Answer 1

The molecular weight of NH4Cl is (14.01 + 4*1.008 + 35.45) = 53.492, or 53.492 g/mole
0.04*53.492*(250/1,000) = 534.92 mg
Dissolve 535 mg in 250 ml H2O.