CHEM 120 Week 3 Solution Chemistry Calculations

CHEM 120 Week 3 Solution Chemistry Calculations

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Chamberlain University

CHEM-120 Intro to General, Organic & Biological Chemistry

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CHEM120 Practice – Solution Chemistry

This section presents a set of solution chemistry problems involving calculations of concentration, molarity, dilution, and osmolarity. Each problem is rephrased for clarity and organized in paragraph and table formats where applicable, maintaining APA formatting.

Percent Concentration of Ammonium Hydroxide Solution

To determine the percent concentration by volume (v/v) of a solution, one must divide the volume of solute by the total volume of the solution and multiply by 100. In this case, 22.5 mL of ammonium hydroxide (NH₄OH) is dissolved in 500 mL of solution.

Calculation:

Sodium Hydroxide Solution Analysis

A solution is prepared by dissolving 77.7 grams of sodium hydroxide (NaOH) into 2.5 liters of water. The solute, percent concentration, and molarity are calculated as follows:

a. The solute in this solution is NaOH.

b. The percent concentration (mass/volume) is found by dividing the solute mass by the total volume in milliliters, multiplied by 100.

c. Molarity is determined using the 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}

Iron (III) Chloride Solution Properties

A solution is prepared by dissolving 5.3 grams of FeCl₃ in water, reaching a total volume of 100 mL (0.100 L).

a. To calculate molarity, use the molar mass of FeCl₃ (162.20 g/mol):

Moles of FeCl₃=5.3 g162.20 g/mol=0.033 mol\text{Moles of FeCl₃} = \frac{5.3\text{ g}}{162.20\text{ g/mol}} = 0.033\text{ mol} Molarity=0.033 mol0.100 L=0.32 M\text{Molarity} = \frac{0.033\text{ mol}}{0.100\text{ L}} = 0.32\text{ M}

b. Osmolarity considers ion dissociation. FeCl₃ dissociates into 4 ions (1 Fe³⁺ and 3 Cl⁻):

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

Volume Adjustment for Nitric Acid Dilution

To determine how much of a 1.2 M nitric acid solution can be made using 10 mL of a 14 M stock solution, the dilution formula is applied:

C1V1=C2V2→V2=C1×V1C2=14×101.2=116.7 mL≈117 mLC_1V_1 = C_2V_2 \rightarrow V_2 = \frac{C_1 × V_1}{C_2} = \frac{14 × 10}{1.2} = 116.7\text{ mL} ≈ 117\text{ mL}

Molarity Post Dilution of Sulfuric Acid

When 25 mL of a 3.5 M sulfuric acid (H₂SO₄) solution is diluted to a final volume of 250 mL:

C1V1=C2V2→C2=C1×V1V2=3.5×25250=0.35 MC_1V_1 = C_2V_2 \rightarrow C_2 = \frac{C_1 × V_1}{V_2} = \frac{3.5 × 25}{250} = 0.35\text{ M}

Dilution Effect on Saline Concentration

If 100 mL of a 20% saline solution is diluted to 500 mL, the new concentration is calculated using:

C_1V_1 = C_2V_2 \rightarrow C_2 = \frac{20% × 100}{500} = 4.0\%

Ethanol Solution Preparation

To prepare 1 liter of a 2% ethanol solution using a 25% ethanol stock:

C1V1=C2V2→V1=C2×V2C1=2×100025=80 mLC_1V_1 = C_2V_2 \rightarrow V_1 = \frac{C_2 × V_2}{C_1} = \frac{2 × 1000}{25} = 80\text{ mL}

References

Brown, T. L., LeMay, H. E., Bursten, B. E., Murphy, C., & Woodward, P. (2018). Chemistry: The Central Science (14th ed.). Pearson.

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

CHEM 120 Week 3 Solution Chemistry Calculations

Chang, R., & Goldsby, K. (2016). Chemistry (12th ed.). McGraw-Hill Education.