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Molarity (M) is the concentration unit defined as moles solute Liter of solution

Molarity (M) is the concentration unit defined as moles solute Liter of solution Reminder: the volume of the solution is the volume occupied by the entire mixture of solvent and solute.

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Molarity (M) is the concentration unit defined as moles solute Liter of solution

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  1. Molarity (M) is the concentration unit defined as moles solute Liter of solution Reminder: the volume of the solution is the volume occupied by the entire mixture of solvent and solute. Molarity

  2. Calculations for Molarity problems are handled very much like those already examined for percentage composition. Example: What is the molarity of a glucose solution which contains 10.0 g glucose (C6H12O6) in 100 mL H2O? Molarity

  3. Notice that the FW or MW of glucose is not given in the problem. The first chore is to calculate a value. C6H12O6 FW = (6C x 12.01)+(12H x 1.01)+(6O x 16.0)= 180.18 So, 1 mole = 180.18 g glucose. Molarity

  4. The next thing to determine is how many moles glucose are in 10.0 g? 1 mole ? = 180.18 g 10.0 g ? = 10.0 g x 1 mole = 0.0555 mole 180.18 g Molarity

  5. Now the molarity can be computed: 0.0555 mole = 0.555 mole/L = 0.555 M 0.100 L Notice the conversion 100 mL = 0.100 L is possible easily because 1000 mL = 1.000 L Molarity

  6. Another example How many grams glucose (FW 180) are contained in 250 mL of 0.10 M aqueous glucose? Analysis: grams are not directly obtained from molarity, but can be obtained from the number of moles and the FW. So the question becomes first, Molarity

  7. How many moles glucose are contained in 250 mL 0.10 M glucose solution? Work from the definition of molarity: 0.10 mole ? = 1.0 L 0.250 L ? = (0.10 mole)x(0.250 L)/(1.0 L) = 0.0250 mole Molarity

  8. The final step is to ask and answer, how many grams are in 0.0250 moles glucose? Work from definition of mole 180 g glucose ? 1 mole 0.0250 mole ? = 180 x 0.0250/1 = 4.50 g glucose = Molarity

  9. Dilution Calculations Very often a small amount of a concentrated solution is mixed with more solvent (diluted) to make a larger volume of a solution that has a lower concentration. This is exactly like adding water to a can of frozen juice concentrate to make a pitcher of oj. Molarity

  10. Add 3 cans water oj concentrate Pitcher of oj All the pulp here… Ends up in the pitcher Molarity

  11. Notice that Molarity = moles L soln This means that Molarity x volume = moles solute in solution Because Molarity x L = moles x L = moles L soln Molarity

  12. So going back to the juice example, Molarity oj concentrate x volume = amount of pulp = Molarity of juice x volume juice In general: Mcon x Vcon = Mdil x Vdil M = molarity, V = volume, dil = dilute, con = concentrated Molarity

  13. A sample dilution problem What volume of 3.0 M HCl is needed to make 250.0 mL of 0.75 M HCl? First, identify the data related to the diluted and concentrated solutions. The more concentrated solution has higher Molarity. Here, 250.0 mL is the volume of the dilute solution, the 0.75 M HCl. Concentrated = 3.0 M HCl Molarity

  14. Thus, Mcon x Vcon = Mdil x Vdil has Mcon = 3.0 M, Vcon = ?, Mdil = 0.75 M, and Vdil = 250.0 mL. Any unit of volume can be used as long as it is the same on both sides of the equation. So we have (3.0 M) x ? = (0.75M)x(250.0 mL). ? = (0.75 x 250.0)/3.0 = 62.5 mL Molarity

  15. Reality check: the volume of the concentrated solution should always be smaller than the volume of the dilute solution. Think about it…. Molarity

  16. Another sample problem What is the molarity of the solution the results by mixing 50.0 mL 16 M HNO3 with enough water to give a final volume of 2.0 L? “Dilute” => ? Molarity & 2.0 L “Concentrated” => 5.0 mL and 16 M Notice: volumes aren’t same units Molarity

  17. First change volumes to L units: 50.0 mL = 0.050 L Now use Mcon x Vcon = Mdil x Vdil With Mcon = 16 M, Vcon = 0.050 L and Mdil = ? , Vdil = 2.0 L 16 M x 0.050 L = ? X 2.0 L (16 M x 0.050 L)/(2.0 L) = ? = 0.40 M Molarity

  18. The strategy we just have been using works for any type of concentration units and any type of volume units because Concentration x volume = amount of stuff dissolved Same units of volume & same units of concentration must be used on both sides of the equation. Molarity

  19. For example people are often told how to estimate the amount of alcohol consumed by the following relationship: 12 oz beer = 4 oz wine = 1 shot (1.5 oz) distilled liquor This statement is reliable because beer is around 4 % alcohol, wine is about 15 % alcohol, and distilled… Molarity

  20. spirits about 40% alcohol. Thus, Beer: 12 oz x 4.5% = 54 oz% Wine: 4 oz x 15% = 60 oz% Distilled spirits: 1.5 oz x 40% = 60 oz% So, if you are into volume drinking, drink beer…there is a lower concentration of alcohol. Knowing some chemistry can help prevent drunk driving! Molarity

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