What is the molarity of a solution of $KNO_3$ that contains 404 grams of $KNO_3$ in 2.00 liters of solution?

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Answer 1 · 2016-04-21T21:17:32

Approx. $2$ $mol*L^-1$ .

$"Concentration "= "Moles of solute"/"Volume of solution"$

We assume that potassium nitrate dissolves in the given volume of solvent without gross change in volume.

$"Concentration "=(404*cancelg)/(101.10*cancelg*mol^-1)xx(1/(2.00*L))$ $=$ $??mol*L^-1$ .

What is the molarity of a solution of KNO_3KNO_3 that contains 404 grams of KNO_3KNO_3 in 2.00 liters of solution?