General Chemistry Exam 1 Study Guide – Step-by-Step Guidance

Q1. How many grams of NaOH are needed to prepare 5.00 L of 0.250 M NaOH (40.0 g/mol)?

Background

Topic: Solution Stoichiometry (Molarity)

This question tests your ability to use molarity to calculate the mass of solute required to make a solution of a given volume and concentration.

Key Terms and Formulas:

• Molarity (M): $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Moles: $\text{moles} = M \times V$

• Mass: $\text{mass} = \text{moles} \times \text{molar mass}$

Step-by-Step Guidance

1. Calculate the number of moles of NaOH needed using the molarity and volume: $\text{moles} = M \times V$.

2. Plug in the values: $M = 0.250$ M, $V = 5.00$ L.

3. Multiply the number of moles by the molar mass of NaOH to find the mass required: $\text{mass} = \text{moles} \times 40.0$ g/mol.

Try solving on your own before revealing the answer!

Final Answer: 50.0 g

$\text{moles} = 0.250 \times 5.00 = 1.25$

$\text{mass} = 1.25 \times 40.0 = 50.0$ g

So, you need 50.0 grams of NaOH to prepare the solution.

Q2. What is the molality of a KF solution that contains 3.0 g of KF dissolved in 25.0 ml of water (density = 0.975 g·ml-1)?

Background

Topic: Molality Calculation

This question tests your ability to calculate molality, which is the number of moles of solute per kilogram of solvent.

Key Terms and Formulas:

• Molality ($m$): $m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

• Moles: $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$

• Density: $\text{mass of solvent} = \text{density} \times \text{volume}$

Step-by-Step Guidance

1. Calculate the moles of KF: $\text{moles KF} = \frac{3.0}{\text{molar mass of KF}}$.

2. Find the mass of water using density and volume: $\text{mass} = 0.975 \times 25.0$ ml.

3. Convert the mass of water to kilograms.

4. Calculate molality: $m = \frac{\text{moles KF}}{\text{kg water}}$.

Try solving on your own before revealing the answer!

Final Answer: 2.5 m

$\text{moles KF} = \frac{3.0}{58.1} \approx 0.052$

$\text{mass water} = 0.975 \times 25.0 = 24.4$ g $= 0.0244$ kg

$m = \frac{0.052}{0.0244} \approx 2.13$ m

So, the molality is approximately 2.13 m.

Q3. Define the following terms: Strong electrolyte and Weak electrolyte.

Background

Topic: Electrolytes

This question tests your understanding of the difference between strong and weak electrolytes in solution chemistry.

Key Terms:

• Strong electrolyte: A substance that completely dissociates into ions in solution.

• Weak electrolyte: A substance that partially dissociates into ions in solution.

Step-by-Step Guidance

1. Recall that strong electrolytes include most salts, strong acids, and strong bases.

2. Weak electrolytes include weak acids and weak bases, which only partially ionize.

3. Think about examples: NaCl (strong), acetic acid (weak).

Try writing your own definitions before revealing the answer!

Final Answer:

Strong electrolyte: Completely dissociates into ions in solution (e.g., NaCl).

Weak electrolyte: Partially dissociates into ions in solution (e.g., acetic acid).

Q4. Circle all of the water-insoluble compounds from the following list:

NH4OH, H2SO4, CaCO3, acetic acid, potassium nitrate, sodium sulfide, magnesium phosphate, Ba(OH)2, H2O2

Background

Topic: Solubility Rules

This question tests your ability to apply solubility rules to identify water-insoluble compounds.

Key Terms:

• Solubility rules: Guidelines for predicting whether a compound will dissolve in water.

Step-by-Step Guidance

1. Recall common solubility rules: Nitrates, ammonium, and alkali metal salts are generally soluble.

2. Carbonates, phosphates, and sulfides are often insoluble unless paired with ammonium or alkali metals.

3. Examine each compound and apply the rules to determine solubility.

Try identifying the insoluble compounds before revealing the answer!

Final Answer:

CaCO3, magnesium phosphate

These compounds are generally insoluble in water.

Q5. exam

Background

Topic: Rate Laws

This question tests your ability to write the rate law for a reaction based on its order.

Key Terms and Formulas:

• Rate law: $\text{Rate} = k[\text{NO}_2]^n$

• Second order: $n = 2$

Step-by-Step Guidance

1. Identify the reactant(s) involved in the rate law.

2. Since the reaction is second order, the exponent for NO2 is 2.

3. Write the general form: $\text{Rate} = k[\text{NO}_2]^2$.

Try writing the rate law before revealing the answer!

Final Answer:

$\text{Rate} = k[\text{NO}_2]^2$

This is the rate law for a second order reaction with respect to NO2.

Q6. Given the reaction [A] + [B] → [C], and the rate law: Rate = k[A]^m[B]^n, answer the following:

Background

Topic: Rate Law and Reaction Order

This question tests your understanding of reaction order, rate law exponents, and units.

Key Terms:

• Rate law: $\text{Rate} = k[A]^m[B]^n$

• Order: Sum of exponents m and n

Step-by-Step Guidance

1. Recall that the overall order is the sum of the exponents: $m + n$.

2. Use the integrated rate law to determine exponents if data is given.

3. Check the units for [A] and [B] (should be M, not M·s-1).

Try matching the statements to the correct concepts before revealing the answer!

Final Answer:

a) True: 2nd order overall if m + n = 2.

d) True: First order in B, second order in A if n = 1, m = 2.

e) False: [A] and [B] are in units of M.

Q7. Consider the reaction NO(g) + O3(g) ⟶ NO2(g) + O2(g) and the following data:

Determine the rate law using this data including the value of k.

Background

Topic: Determining Rate Law from Experimental Data

This question tests your ability to use experimental data to determine the rate law and rate constant.

Key Terms and Formulas:

• Rate law: $\text{Rate} = k[\text{NO}]^m[\text{O}_3]^n$

• Use data to find m and n by comparing rates and concentrations.

Step-by-Step Guidance

1. Compare trials where only one reactant concentration changes to determine the order with respect to each reactant.

2. Use the ratio of rates and concentrations to solve for exponents m and n.

3. Once the rate law is established, use any trial to solve for k.

Try working through the data before revealing the answer!

Final Answer:

$\text{Rate} = k[\text{NO}][\text{O}_3]$

$k = \frac{\text{Rate}}{[\text{NO}][\text{O}_3]}$

Plug in values from any trial to solve for k.

Q8. Calculate the freezing point constant, Kf, for a water/ethanol solution which has a temperature change of -0.75 °C and is composed of 25.0 g of ethanol dissolved in 75.0 ml water (density = 0.985 g/ml).

Background

Topic: Colligative Properties – Freezing Point Depression

This question tests your ability to use freezing point depression to calculate the freezing point constant, Kf.

Key Terms and Formulas:

• Freezing point depression: $\Delta T_f = K_f \times m$

• Molality: $m = \frac{\text{moles solute}}{\text{kg solvent}}$

Step-by-Step Guidance

1. Calculate the moles of ethanol: $\text{moles} = \frac{25.0}{\text{molar mass of ethanol}}$.

2. Find the mass of water in kg: $75.0 \times 0.985$ g $= \text{mass}$, then convert to kg.

3. Calculate molality: $m = \frac{\text{moles ethanol}}{\text{kg water}}$.

4. Use $\Delta T_f = K_f \times m$ to solve for $K_f$.

Try calculating Kf before revealing the answer!

Final Answer:

$K_f = \frac{\Delta T_f}{m}$

Plug in the values to find Kf.

Q9. What is the van’t Hoff factor for 2.50 moles of sulfuric acid dissolved in 200 ml of water?

Background

Topic: Colligative Properties – van’t Hoff Factor

This question tests your understanding of the van’t Hoff factor, which is the number of particles a solute produces in solution.

Key Terms:

• van’t Hoff factor ($i$): Number of particles produced per formula unit.

• For $\text{H}_2\text{SO}_4$, consider its dissociation in water.

Step-by-Step Guidance

1. Write the dissociation equation for sulfuric acid in water.

2. Count the total number of ions produced per formula unit.

3. van’t Hoff factor is the sum of all ions.

Try determining the van’t Hoff factor before revealing the answer!

Final Answer:

For $\text{H}_2\text{SO}_4$, $i = 3$ (if fully dissociated).

Q10. What is the van’t Hoff factor for an unknown solution with a freezing point constant of 1.86°C/m and a concentration of 1.77 m that shows a temperature drop of 6.58 °C?

Background

Topic: Colligative Properties – van’t Hoff Factor

This question tests your ability to use freezing point depression to determine the van’t Hoff factor.

Key Terms and Formulas:

• Freezing point depression: $\Delta T_f = i \times K_f \times m$

Step-by-Step Guidance

1. Write the formula: $\Delta T_f = i \times K_f \times m$.

2. Plug in the values for $\Delta T_f$, $K_f$, and $m$.

3. Rearrange to solve for $i$: $i = \frac{\Delta T_f}{K_f \times m}$.

Try calculating the van’t Hoff factor before revealing the answer!

Final Answer:

$i = \frac{6.58}{1.86 \times 1.77}$

Calculate to find the van’t Hoff factor.

Q11. A water-soluble inorganic salt is added to water. It is determined that the salt has aluminum for the cation in a +3 oxidation state. If we have a 0.015 m solution and the Kf for water is 1.86 °C/m, which of the following could be the identity of the anion if the solution’s freezing point is determined to drop by 0.1395 °C?

NO3-, PO43-, SO42-

Background

Topic: Colligative Properties – van’t Hoff Factor and Freezing Point Depression

This question tests your ability to use freezing point depression and the van’t Hoff factor to deduce the formula of a salt.

Key Terms and Formulas:

• Freezing point depression: $\Delta T_f = i \times K_f \times m$

• van’t Hoff factor ($i$): Number of ions produced per formula unit.

Step-by-Step Guidance

1. Write the formula: $\Delta T_f = i \times K_f \times m$.

2. Plug in the values for $\Delta T_f$, $K_f$, and $m$.

3. Calculate $i$ and compare to possible salt formulas (Al(NO3)3, Al2(SO4)3, AlPO4).

Try deducing the anion before revealing the answer!

Final Answer:

Calculate $i$ and match to the salt with the correct number of ions.

Q12. Determine the overall balanced reaction for the following multi-step reaction and identify the intermediate(s):

1) NO2 + NO2 → NO3 + NO 2) NO3 + CO → NO2 + CO2

Background

Topic: Multi-Step Reactions and Intermediates

This question tests your ability to combine elementary steps and identify intermediates.

Key Terms:

• Intermediate: A species produced in one step and consumed in another.

• Overall reaction: Sum of all steps, cancel intermediates.

Step-by-Step Guidance

1. Write both reactions and add them together.

2. Identify and cancel intermediates (species not present in overall reaction).

3. Write the overall balanced equation.

Try combining the steps before revealing the answer!

Final Answer:

Overall: NO2 + NO2 + CO → NO + NO2 + CO2

Intermediate: NO3

Q13. Given the reaction N2(g) + 3H2(g) → 2NH3(g), if the rate of formation of NH3(g) is 2.0 mol/(L·s), what is the rate of consumption of H2(g) in mol/(L·s)?

Background

Topic: Reaction Stoichiometry and Rates

This question tests your ability to relate rates of consumption and formation using stoichiometry.

Key Terms and Formulas:

• Rate relationship: $\frac{1}{2} \frac{d[\text{NH}_3]}{dt} = -\frac{1}{3} \frac{d[\text{H}_2]}{dt}$

Step-by-Step Guidance

1. Write the stoichiometric relationship between NH3 and H2.

2. Set up the rate equation using the coefficients.

3. Plug in the rate of NH3 formation to solve for the rate of H2 consumption.

Try setting up the calculation before revealing the answer!

Final Answer:

Rate of H2 consumption: $\text{Rate} = -\frac{3}{2} \times 2.0$ mol/(L·s)

Calculate to find the rate.