13.3 Units Of Concentration

Example 4

Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL.

1. What is its molarity?
2. What is its mole fraction?

Given: mass of substance and mass and density of solution

Asked for: molarity and mole fraction

Strategy:

A Calculate the number of moles of acetic acid in the sample. Then calculate the number of liters of solution from its mass and density. Use these results to determine the molarity of the solution.

B Determine the mass of the water in the sample and calculate the number of moles of water. Then determine the mole fraction of acetic acid by dividing the number of moles of acetic acid by the total number of moles of substances in the sample.

Solution:

1. A The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass. The volume of the solution equals its mass divided by its density. The calculations follow:

   moles CH 3 CO 2 H = 3.78   g  CH 3 CO 2 H 60.05   g /mol = 0.0629  mol volume = mass density = 100.0   g  solution 1.00   g /mL = 100  mL molarity of CH 3 CO 2 H = moles CH 3 CO 2 H liter solution = 0.0629  mol CH 3 CO 2 H ( 100   mL ) ( 1  L/1000  mL ) = 0.629  M CH 3 CO 2 H

   This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than 12 mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL.

2. B To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have

   moles H 2 O = 96.2   g  H 2 O 18.02   g /mol = 5.34  mol H 2 O

   The mole fraction X of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present:

   X CH 3 CO 2 H = moles  CH 3 CO 2 H moles  CH 3 CO 2 H  +  moles  H 2 O = 0.0629  mol 0.0629  mol + 5. 34 mol = 0.0116 = 1.16 × 10 − 2

   This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01.

Exercise

A solution of HCl gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of HCl per 100.0 g of solution, and its density is 1.10 g/mL.

1. What is its molarity?
2. What is its mole fraction?

Answer:

1. 6.10 M HCl
2. XHCl = 0.111