CHEM 120 Week 3 Solution Chemistry Calculations

CHEM 120 Week 3 Solution Chemistry Calculations

Name

Chamberlain University

CHEM-120 Intro to General, Organic & Biological Chemistry

Prof. Name

Date

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

The percent concentration is calculated using the formula:

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

Answer: NaOH

b. What is the percent concentration of this solution?

The percent concentration is given by:

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?

Molar mass of NaOH = 40.00 g/mol

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 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?

Molar mass of FeCl₃ = 162.20 g/mol, Volume = 0.100 L

moles of solute=5.3 g162.20 g/mol=0.033 mol\text{moles of solute} = \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 into 4 ions (1 Fe³⁺ and 3 Cl⁻).

Osmolarity=0.32 M×4=1.28 osmol/L\text{Osmolarity} = 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 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 C1V1=C2V2C_1V_1 = C_2V_2:

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.

CHEM 120 Week 3 Solution Chemistry Calculations

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