Question 1: Calculate the percent concentration of a 500 mL solution made by dissolving 22.5 mL of NH₄OH in water

The percent concentration of a solution represents the ratio of the volume of solute to the total volume of the solution, multiplied by 100. For this solution, the calculation is based on the volume/volume percent formula:

v/v %=volume of solutevolume of solution×100\text{v/v \%} = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100

Substituting the given values:

v/v %=22.5 mL500 mL×100=4.50%\text{v/v \%} = \frac{22.5 \, \text{mL}}{500 \, \text{mL}} \times 100 = 4.50\%

Answer: 4.50%

Question 2: A 2.5 liter solution is prepared by dissolving 77.7 grams of NaOH in water

a. What is the solute in this solution?

The solute in this solution is NaOH, which is dissolved in water to form the solution.

b. What is the percent concentration of this solution?

The percent concentration in mass/volume terms is calculated as:

m/v %=mass of solute (g)volume of solution (mL)×100\text{m/v \%} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 m/v %=77.7 g2500 mL×100=3.11%\text{m/v \%} = \frac{77.7 \, g}{2500 \, mL} \times 100 = 3.11\%

Answer: 3.11%

c. What is the molarity of this solution?

The molarity is determined by dividing the moles of solute by the volume of solution in liters. First, calculate the moles of NaOH:

Moles of solute=77.7 g40.00 g/mol=1.94 mol\text{Moles of solute} = \frac{77.7 \, g}{40.00 \, g/mol} = 1.94 \, mol

Then, the molarity is:

Molarity=1.94 mol2.5 L=0.777 M\text{Molarity} = \frac{1.94 \, mol}{2.5 \, L} = 0.777 \, M

Answer: 0.777 M NaOH

Question 3: A solution is prepared by dissolving 5.3 g of FeCl₃ in water to a total volume of 100 mL

a. What is the molarity of this solution?

The molarity is calculated using the formula:

Molarity=moles of solutevolume of solution (L)\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution (L)}} Moles of FeCl₃=5.3 g162.20 g/mol=0.033 mol\text{Moles of FeCl₃} = \frac{5.3 \, g}{162.20 \, g/mol} = 0.033 \, mol Molarity=0.033 mol0.100 L=0.32 M\text{Molarity} = \frac{0.033 \, mol}{0.100 \, L} = 0.32 \, M

Answer: 0.32 M FeCl₃

b. What is the osmolarity of this solution?

FeCl₃ dissociates in water into four ions: one Fe³⁺ ion and three Cl⁻ ions. Osmolarity is calculated as:

Osmolarity=Molarity×number of ions=0.32 M×4=1.28 osmol/L\text{Osmolarity} = \text{Molarity} \times \text{number of ions} = 0.32 \, M \times 4 = 1.28 \, osmol/L

Answer: 1.28 osmol/L

Question 4: How many mL of a 1.2 M HNO₃ solution can you prepare using 10 mL of a 14 M HNO₃ solution?

Using the dilution formula C1V1=C2V2C_1V_1 = C_2V_2:

V2=C1×V1C2=14 M×10 mL1.2 M=117 mLV_2 = \frac{C_1 \times V_1}{C_2} = \frac{14 \, M \times 10 \, mL}{1.2 \, M} = 117 \, mL

Answer: 117 mL

Question 5: What is the molarity of a solution prepared by diluting 25 mL of a 3.5 M H₂SO₄ solution to a final volume of 250 mL?

Using the dilution relationship C1V1=C2V2C_1V_1 = C_2V_2:

C2=C1×V1V2=3.5 M×25 mL250 mL=0.35 MC_2 = \frac{C_1 \times V_1}{V_2} = \frac{3.5 \, M \times 25 \, mL}{250 \, mL} = 0.35 \, M

Answer: 0.35 M

Question 6: The volume of a 20% saline solution is increased from 100 mL to 500 mL. What is the new concentration of this solution?

Using the dilution formula:

C2=C1×V1V2=20%×100 mL500 mL=4.0%C_2 = \frac{C_1 \times V_1}{V_2} = \frac{20\% \times 100 \, mL}{500 \, mL} = 4.0\%

Answer: 4.0%

Question 7: How many mL of a 25% ethanol solution would you require to prepare 1 L of a 2% ethanol solution?

Using C1V1=C2V2C_1V_1 = C_2V_2:

V1=C2×V2C1=2%×1000 mL25%=80 mLV_1 = \frac{C_2 \times V_2}{C_1} = \frac{2\% \times 1000 \, mL}{25\%} = 80 \, mL

Answer: 80 mL

Summary Table

References

Brown, T. L., LeMay, H. E., Bursten, B. E., & Murphy, C. J. (2014). Chemistry: The Central Science (13th ed.). Pearson.

Zumdahl, S. S., & Zumdahl, S. A. (2017). Chemistry (10th ed.). Cengage Learning.

CHEM 120 Week 3 Solution Chemistry Calculations.