# Hi! I can't seem to figure out how to do this problem. If anybody could help me, that would be great!?

A 2.200×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.1 mL . The density of water at 20.0∘C is 0.9982 g/mL.

a) Calculate the concentration of the glycerol solution in percent by mass.

b)Calculate the concentration of the glycerol solution in parts per million.

### 2 Answers

- hcbiochemLv 71 month ago
a) Mass glycerol = 2.200X10^-2 mol X 92.09 g = 2.026 g glycerol

mass water = 999.1 mL X 0.9982 g/mL = 997.3016 g water

Mass% glycerol = (2.026 g / (2.03 + 997.302 g) X 100 = 0.2027%

b) Parts per million can be interpreted in different ways. One very common way is a mass/volume meaning milligrams of solute / L of solution.

22.00 mg / L = 22.00 ppm

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- Anonymous1 month ago
You have 1 L of solution so you have .200×10−2 mole glycerol. What is the mass of glycerol?

Assuming the density of the solution = the density of water then 1 L of solution has a mass of 998.2 g.

mass % glycerol = mass of glycerol / 998.2 g * 100

ppm glycerol by mass = mass of glycerol / 998.2 g * 1 * 10^6

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