CHEM 120 Week 3 Solution Chemistry Calculations

Grace J.
August 13, 2025August 13, 2025
CHEM-120, CU BSN

Student Name

Chamberlain University

CHEM-120 Intro to General, Organic & Biological Chemistry

Prof. Name

Date

CHEM120 Practice – Solution Chemistry

This section presents a series of solution chemistry problems, each demonstrating practical applications of concentration calculations, molarity, dilutions, and osmolarity concepts. The solutions are calculated step-by-step with explanations for clarity. Where applicable, the results are presented in both paragraph explanations and table format for easy reference.

1. Percent Concentration of an NH₄OH Solution

Question:
Calculate the percent by volume (% v/v) of a solution prepared by dissolving 22.5 mL of ammonium hydroxide (NH₄OH) into enough water to make 500 mL of solution.

Answer:
The percent by volume is determined using the equation:

%v/v=Total volume of solutionVolume of solute×100

Substituting values: %v/v=22.5 mL500 mL×100=4.50%\% v/v = \frac{22.5 \text{ mL}}{500 \text{ mL}} \times 100 = 4.50\%%v/v=500 mL22.5 mL×100=4.50%

Result: The concentration is 4.50% v/v NH₄OH.

2. Properties of a Sodium Hydroxide Solution

Question:
A 2.5 L solution is prepared by dissolving 77.7 g of sodium hydroxide (NaOH) in water.
a) Identify the solute.
b) Calculate the mass/volume percent concentration.
c) Calculate the molarity of the solution.

Answer:

a) Solute Identification

The solute in this case is NaOH (solid sodium hydroxide).

b) Percent Concentration (m/v %)

Formula:

%m/v=Mass of solute (g)Volume of solution (mL)×100\% m/v = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100

Substitution:

%m/v=77.7 g2500 mL×100=3.11%\% m/v = \frac{77.7 \text{ g}}{2500 \text{ mL}} \times 100 = 3.11\%

c) Molarity

Molar mass of NaOH = 40.00 g/mol

Moles of NaOH=77.7 g40.00 g/mol=1.94 mol\text{Moles of NaOH} = \frac{77.7 \text{ g}}{40.00 \text{ g/mol}} = 1.94 \text{ mol} Molarity=1.94 mol2.5 L=0.777 M\text{Molarity} = \frac{1.94 \text{ mol}}{2.5 \text{ L}} = 0.777 \text{ M}

Summary Table:

Property	Value
Solute	NaOH
Percent Concentration (m/v %)	3.11%
Molarity	0.777 M

3. Molarity and Osmolarity of an FeCl₃ Solution

Question:
A solution is prepared by dissolving 5.3 g of ferric chloride (FeCl₃) into enough water to make 100 mL of solution.
a) Calculate the molarity.
b) Calculate the osmolarity.

Answer:

a) Molarity

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

Moles of FeCl₃=5.3 g162.20 g/mol=0.0327 mol\text{Moles of FeCl₃} = \frac{5.3 \text{ g}}{162.20 \text{ g/mol}} = 0.0327 \text{ mol} Molarity=0.0327 mol0.100 L=0.327 M\text{Molarity} = \frac{0.0327 \text{ mol}}{0.100 \text{ L}} = 0.327 \text{ M}

Rounded to 0.32 M FeCl₃.

b) Osmolarity

Since FeCl₃ dissociates into 1 Fe³⁺ ion and 3 Cl⁻ ions (total of 4 particles):

Osmolarity=0.32 M×4=1.28 osmol/L\text{Osmolarity} = 0.32 \text{ M} \times 4 = 1.28 \text{ osmol/L}

Summary Table:

Property	Value
Molarity	0.32 M
Osmolarity	1.28 osmol/L

4. Dilution of HNO₃ Solution

Question:
How many milliliters of a 1.2 M nitric acid (HNO₃) solution can be made from 10 mL of a 14 M HNO₃ stock solution?

Answer:

Using the dilution equation:

C1V1=C2V2C_1V_1 = C_2V_2 V2=C1×V1C2=14 M×10 mL1.2 M=116.67 mLV_2 = \frac{C_1 \times V_1}{C_2} = \frac{14 \text{ M} \times 10 \text{ mL}}{1.2 \text{ M}} = 116.67 \text{ mL}

Result: Approximately 117 mL of 1.2 M HNO₃ can be prepared.

5. Molarity After Dilution of H₂SO₄

Question:
What is the molarity of a sulfuric acid (H₂SO₄) solution prepared by diluting 25 mL of a 3.5 M stock solution to a final volume of 250 mL?

Answer:

C1V1=C2V2C_1V_1 = C_2V_2 C2=3.5 M×25 mL250 mL=0.35 MC_2 = \frac{3.5 \text{ M} \times 25 \text{ mL}}{250 \text{ mL}} = 0.35 \text{ M}

Result: The diluted solution has a molarity of 0.35 M H₂SO₄.

6. Concentration After Volume Increase (Saline Solution)

Question:
A 20% saline solution (100 mL) is diluted to a final volume of 500 mL. What is the new concentration?

Answer:

C2=20%×100 mL500 mL=4.0%C_2 = \frac{20\% \times 100 \text{ mL}}{500 \text{ mL}} = 4.0\%

Result: The new concentration is 4.0% saline.

7. Preparing a Diluted Ethanol Solution

Question:
How many milliliters of a 25% ethanol solution are needed to prepare 1 L of a 2% ethanol solution?

Answer:

C1V1=C2V2C_1V_1 = C_2V_2 V1=2%×1 L25%=0.080 L=80 mLV_1 = \frac{2\% \times 1 \text{ L}}{25\%} = 0.080 \text{ L} = 80 \text{ mL}

Result: 80 mL of 25% ethanol is required.

References

Brown, T. L., LeMay, H. E., Bursten, B. E., Murphy, C., Woodward, P., & Stoltzfus, M. W. (2021). Chemistry: The central science (15th ed.). Pearson.

Zumdahl, S. S., & Zumdahl, S. A. (2020). Chemistry (11th ed.). Cengage Learning.