How many grams of potassium chloride (MW K = 39, MW Cl = 35.5) are needed to make 125 mL of a solution containing 200 mOsmol/L? Do not include units. Round to the nearest hundredth.

Question

How many grams of potassium chloride (MW K = 39, MW Cl = 35.5) are needed to make 125 mL of a solution containing 200 mOsmol/L? Do not include units. Round to the nearest hundredth.

Answer 1

KCl ==> K⁺ + Cl^- so 1 mole of KCl = 2 osmoles (produces 2 particles)

125 ml of solution = 0.125 L of solution

200 mOsmol / L x 0.125 L = 25 mOsmol required

25 mOsmols x 1 mmol KCl / 2 mosmols = 12.5 mmoles KCl required

molar mass KCl = 39 + 35.5 = 74.5 mg / mmole

12.5 moles KCl x 74.5 mg / mmole = 931 mg = 0.931 g = 0.93 (rounded to hundredths w/o units)