1. Other units of concentration

2. Dilution Equation MV = MV Molarity (volume) before dilution
= molarity (volume) after dilution
How many liters of 12 M H2SO4 do you need to make 1.2 L of .75 M?
12 M (V) = .75 M (1.2 L)
V = .075 L (75 mL)

3. Mass Percent Mass percent = mass of solute x 100 mass of solution
Or = grams of solute x 100
grams of solute + grams of solvent

4. Problem
35 g of NaCl is dissolved in 115 g of water, what is the mass percent? What is the molarity if the final solution has a density of 1.1 g/mL?
35 g / (35g + 115g) x 100
23 %
35 g x 1 mol/ g= mol NaCl
150 g x 1mL / 1.1 g = mL = L
M = mol / L
M = 4.4 M

5. Convert
Convert 1.2 M CuSO4 solution to mass percent, if the solution has a density 1.1 g/mL.
1.2 M= 1.2 mol CuSO4 / 1 L solution
1.2 mol x g / 1 mol = g
1 L = 1000 mL x 1.1 g/1 mL = 1100 g of solution
Mass percent = g / 1100 g x100 = 17 %

6. More practice
55 g of CaCl2 is dissolved in 115 g of water, what is the mass percent? What is the molarity if the final solution has a density of 1.1 g/mL?

8. More
Convert 1.9 M Ca(NO3)2 solution to mass percent, if the solution has a density 1.3 g/mL.

10. Parts per thousand/million etc.
Mass percent can also be called parts per hundred (although it never is)
Parts per thousand is the same as mass percent except instead of multiplying by a 100 you multiply by 1000.
Parts per million is multiplied by 1,000,000
Pollen counts are normally reported in this

11. Mole Fraction

12. Mole Fraction (these calculations are not in your book)
Mole Fraction or molar fraction is represented by the Greek letter chi, 
 = mole solute/mole solution

13. Convert
Convert a 2.3 M of water solution of aluminum nitrate Al(NO3)3 to mole fraction of aluminum nitrate if the density of the solution is 1.14 g/mL.
2.3 M = 2.3 mol Al(NO3)3 /1 L solution
2.3 mol x g/ 1 mol = g Al(NO3)3
1 L = 1000 mL x 1.14 g/1 mL = 1140 g solution
1140 g g = g water
g x 1 mol / g = mol
 = 2.3 mol/ ( ) mol = .060

14. Another
Convert a 1.4 M solution of strontium sulfate, SrSO4, in water to mole fraction if the density of the solution is 1.24 g/mL.
1.4 M = 1.4 mol SrSO4 /1 L solution
1.4 mol x g/ 1 mol = g SrSO4
1 L = 1000 mL x 1.24 g/1 mL = 1240 g solution
1240 g – g = g water
g x 1 mol / g = mol
 = 1.4 mol/ ( ) mol = .025