Molarity Problems


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Example #1: A student placed 11.0 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0 mL sample of this glucose solution was diluted to 0.500L. How many grams of glucose are in 100. mL of the final solution?

Solution path #1:

1) Calculate molarity of first solution (produced by dissolving 11.0 g of glucose):

MV = grams / molar mass

(x) (0.100 L) = 11.0 g / 180.155 g/mol

x = 0.610585 mol/L (I'll carry a few guard digits.)

2) Calculate molarity of second solution (produced by diluting the first solution):

M1V1 = M2V2

(0.0200 L) (0.610585 mol/L) = (0.500 L) (x)

x = 0.0244234 mol/L

3) Determine grams of glucose in 100. mL of second solution:

MV = grams / molar mass

(0.0244234 mol/L) (0.100 L) = x / 180.155 g/mol

x = 0.44 g

Solution path #2:

1) Calculate how much glucose you have in 20.0 mL of the first solution.

11.0 g is to 100. mL as x is to 20.0 mL

Cross-multiply and divide

100x = 11.0 times 20.0

x = 2.2 g

2) When you dilute the 20.0 mL sample to 500.0 mL, you have 2.2 g glucose in the solution.
2.2 g is to 500. mL as x is to 100. mL

Cross-multiply and divide

500x = 2.2 times 100

x = 0.44 g

In 100 mL of the final solution are 0.44 g glucose.


Example #2: Commercial bleach solution contains 5.25% (by mass) of NaClO in water. It has a density of 1.08 g/mL. Caculate the molarity of this solution. (Hints: assume you have 1.00 L of solution; molar mass of NaClO 74.4 g/mol)

Solution:

1) Determine mass of 1.00 L of solution:

(1.08 g/mL) (1000 mL) = 1080 g

2) Determine mass of NaClO in 1080 g of solution:

(1080 g) (0.0525) = 56.7 g

3) Determine moles of NaClO:

56.7 g / 74.4 g/mol = 0.762 mol

4) Determine molarity of solution:

0.762 mol / 1.00 L = 0.762 M

Example #3: Someday!