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# Help with Final Concentration?

Could someone work the steps and provide an answer for these final concentration problems? Thanks!

1. Water added to 12.0 mL of a 2.50 M KNO3 solution gives a volume of 0.150 L.

2. A 5.80 mL sample of a 16.0 M sucrose solution is diluted with water to 110 mL.

3. Water is added to 0.200 L of a 10.4 M {\rm KBr} solution to give a volume of 0.880 L.

### 1 Answer

- 9 years agoFavorite Answer
Simple ratio. Substance contained in one volume is diluted to a final volume. When doing the ratio, the units of volume have to match in the numerator & denominator of the ratio (Liters to liters, milliliters cancel milliliters). I included conversion factors (in #2 it's redundant but showing just so you can see) and labeled the volumes 'init' and 'final' to make clear which is which. See that the units in the numerator & denominator cancel leaving only those you need. This naming of the units & carrying them through computations is standard procedure & helps you recognize screw-ups early.

1. 2.50 mols / L.init KNO3 x 12.0 mL.init x [ 0.001 L/mL ] / 0.150 L.final = 0.200 mols/L.final KNO3

2. 16.0 mols / L.init sucrose x 5.80 mL.init x [ 0.001 L/mL ] / (110 mL.final x [0.001 L/mL] = 0.844 M sucrose

3. 10.4 mols / L.init KBr x 0.200 L.init / 0.880 L.final = 2.36 M KBr