Question

How many grams of potassium chloride (KCI) are needed to prepare 0.750 L of a 1.50 M solution of potassium chloride in water?​

Answer 1

KCl: 74.5g/mol

Molarity = mol/Liters
Since you are given 0.750L and 1.50M, we can plug it into the molarity formula and solve for moles.

1.50M = x mol/0.750L
X = 1.125 moles of KCl needed

You asked for grams so let’s convert the moles of KCl to grams:

1.125 mol x (74.5g/mol) = 83.8g KCl