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?
1 answer:
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
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