Asked by jen

how many grams of potassium nitrate are needed to prepare 250.0ml of a 0.03M potassium nitrate solution?
Answered by Damon
.03 / 4 = .0075 mols of KNO3 needed

.0075 * 101 g/mol = .758 grams
Answered by DrBob222
How many mols do you want? That's mols = M x L = 0.03 x 0.250 = 0.0075 mols. Then
0.0075 mols x 101 g/mol= ?
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