Login or Register

800.334.5551 Live Chat (offline)

Chemistry Solutions Practice Problems

1. Molar solutions

a. Describe how you would prepare 1 L of a 1 M solution of sodium chloride. The gram formula weight of sodium chloride is 58.44 g/mol.

Answer: To make a 1 M solution of sodium chloride, dissolve 58.44 g sodium chloride in 500 mL water in a 1000-mL volumetric flask. When all the solid is dissolved and the solution is at room temperature, dilute to the mark and invert the flask several times to mix.

b. Describe how you would prepare 1 L of a 2 M solution of acetic acid. The gram formula weight of acetic acid is 60.05 g/mol, and its density is 1.049 g/mL.

Answer: To make a 2 M solution of acetic acid, dissolve 120.1 g acetic acid in 500 mL distilled or deionized water in a 1000-mL volumetric flask. Since acetic acid is a liquid, it may also be measured by volume. Divide the mass of acid by its density (1.049 g/mL) to determine the volume (114 mL). Use either 120.1 g or 114 mL acetic acid to make the solution. Swirl the flask gently to mix the solution. Once the solution is at room temperature, dilute to the mark and invert the flask several times to mix.

2. Percent solutions

a. Describe how you would prepare 100 g of a solution that is 0.5% phenolphthalein by mass.

Answer: Since the solute (phenolphthalein) is a solid, the solution is percent by mass. Mass percent means the number of grams of solute per 100 g of solution.   

mass percent = (mass of solute/mass of solution) × 100%
 
mass of solute = mass percent × mass of solution/100%
 
= 0.5% x 100 g/100%
 
= 0.5 g

Since the total mass of the solution equals 100 g, the remaining 99.5 g of the solution is water. To prepare the solution, dissolve 0.5 g phenolphthalein in 99.5 g distilled or deionized water.

b.<> Describe how you would prepare 100 mL of a solution that is 22% acetic acid by volume.

Answer: Volume percent means the number of milliliters of solute per 100 mL of solution. Dilute 22 mL acetic acid with distilled or deionized water to make 100 mL of solution.

3. Dilutions

a. Describe how you would prepare 1.0 L of a 0.10 M solution of sulfuric acid from a 3.0 M solution of sulfuric acid.

Answer: Calculate the volume of 3.0 M sulfuric acid needed to prepare the dilution.  

Mreagent × Vreagent = Mdilution × Vdilution
  
3.0 M × Vreagent = 0.10 M × 1.0 L
 
Vreagent = 0.033 L = 33 mL

Slowly add 33 mL of 3.0 M sulfuric acid to a 1000-mL volumetric flask half-filled with distilled or deionized water and swirl the flask to mix. Once the solution is at room temperature, dilute to the mark with water and invert the flask several times to mix.

b.Describe how you would prepare 500 mL of a 0.25 M solution of sodium hydroxide from a 5.0 M solution of sodium hydroxide.

Answer: Calculate the volume of 5.0 M sodium hydroxide needed to prepare the dilution.

Mreagent × Vreagent = Mdilution × Vdilution
 
5.0 M × Vreagent = 0.25 M × 0.500 L 
 
Vreagent = 0.025 L = 25 mL

Slowly add 25 mL of 5.0 M sodium hydroxide to a 500-mL volumetric flask half-filled with distilled or deionized water and swirl the flask to mix. Once the solution is at room temperature, dilute to the mark with water and invert the flask several times to mix.

4. Special cases

a. Describe how you would prepare 500 mL of a 1.0 M solution of potassium chloride that is 93.0% pure. The gram formula weight of potassium chloride is 74.56 g/mol.

Answer: Calculate the mass of impure potassium chloride needed.

mass of impure potassium chloride = M pure × V pure × gram formula weight / percent purity
 
= 1.0 M × 0.500 L × 74.56 g/mol
0.930

= 40 g

Slowly add 40 g of 93% potassium chloride to a 500-mL volumetric flask half-filled with distilled or deionized water and swirl the flask to mix. When all the solid is dissolved and the solution is at room temperature, dilute to the mark and invert the flask several times to mix.

b. Describe how you would prepare 500 mL of a 1.0 M solution of phosphoric acid from 85.0% phosphoric acid that is pure. The gram formula weight of phosphoric acid is 98.00 g/mol, and the density of 85.0% phosphoric acid is 1.685 g/mL.

Answer: Calculate the volume of 85.0% phosphoric acid needed.

volume of impure phosphoric acid = M pure × V pure × gram formula weight / (percent purity x density) 

= (1.0 M × 0.500 L x 98.00 g/mol) ÷ 1.685 g/mL
0.850

= 34 mL

Slowly add 34 mL of 85.0% phosphoric acid to a 500-mL volumetric flask half-filled with distilled or deionized water and swirl the flask to mix. Once the solution is at room temperature, dilute to the mark with water and invert the flask several times to mix.

5. Normal solutions

Describe how you would prepare 1000 mL of a 1.0 N solution of magnesium hydroxide. The gram formula weight of magnesium hydroxide is 58.33 g/mol.

Answer: Magnesium hydroxide contains two hydroxyl groups. One-half mole of magnesium hydroxide, therefore, accepts one mole of protons. To prepare a 1.0 N solution of magnesium hydroxide, slowly add 29 g magnesium hydroxide to a 500-mL volumetric flask half-filled with distilled or deionized water and swirl the flask to mix.  When all the solid is dissolved and the solution is at room temperature, dilute to the mark and invert the flask several times to mix. 

Carolina Biological Supply

© Carolina Biological Supply Company

2700 York Road, Burlington, NC 27215-3398 • 800.334.5551