2004 Ap Chem Frq Form B Solutions And Analysis

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AP Chemistry FRQ Form B Solutions and Analysis

The AP Chemistry FRQ (Free Response Questions) Form B is a critical component of the Advanced Placement Chemistry exam. It assesses students' ability to think critically, solve problems, and apply chemical concepts to real-world scenarios. In this article, we will provide solutions and analysis for the 2004 AP Chemistry FRQ Form B.

Section A: Questions 1-3

Question 1: A sample of oxygen gas (O2) is collected over water at a temperature of 25°C and a pressure of 1 atm. The partial pressure of oxygen in the sample is 0.8 atm.

(a) Calculate the mole fraction of oxygen in the sample.

(b) Calculate the mass of oxygen in 1.0 L of the sample at STP (Standard Temperature and Pressure).

Solution 1a:

(a) Mole fraction of oxygen = Partial pressure of oxygen / Total pressure Mole fraction of oxygen = 0.8 atm / 1 atm = 0.8

Solution 1b:

(b) Mass of oxygen in 1.0 L of the sample at STP: Use the ideal gas law: PV = nRT Rearrange the equation to solve for n: n = PV / RT n = (1 atm × 1 L) / (0.08206 L atm/mol K × 273 K) = 0.0446 mol Mass of oxygen = Moles of oxygen × Molar mass of oxygen Mass of oxygen = 0.0446 mol × 32 g/mol = 1.43 g

Section B: Questions 4-6

Question 4: A solution of potassium carbonate (K2CO3) is prepared by dissolving 10.0 g of the solid in 100 mL of water. The resulting solution has a pH of 11.5.

(a) Calculate the molar concentration of hydroxide ions (OH-) in the solution.

(b) Calculate the concentration of carbonate ions (CO32-) in the solution.

Solution 4a:

(a) pH = -log[H+] [H+] = 10^(-pH) = 10^(-11.5) = 3.16 × 10^(-12) M Kw = [H+][OH-] = 1.0 × 10^(-14) [OH-] = Kw / [H+] = 1.0 × 10^(-14) / 3.16 × 10^(-12) = 3.16 × 10^(-3) M

Solution 4b:

(b) Concentration of carbonate ions (CO32-): Use the equation: K2CO3 → 2K+ + CO32- Molar concentration of K2CO3 = Moles of K2CO3 / Volume of solution (L) Molar concentration of K2CO3 = 10.0 g / 100 mL × (1 mol / 138 g) = 0.0725 M Concentration of carbonate ions = Molar concentration of K2CO3 = 0.0725 M

Section C: Questions 7-9

Question 7: A sample of nitrogen gas (N2) is heated from 25°C to 100°C at constant pressure.

(a) Calculate the percentage increase in the volume of the gas.

(b) Calculate the change in the number of moles of gas.

Solution 7a:

(a) Use Charles' Law: V1 / T1 = V2 / T2 V1 / 298 K = V2 / 373 K V2 = V1 × (373 K / 298 K) = 1.25V1 Percentage increase in volume = ((V2 - V1) / V1) × 100% Percentage increase in volume = ((1.25V1 - V1) / V1) × 100% = 25%

Solution 7b:

(b) Change in the number of moles of gas: Use the ideal gas law: PV = nRT Rearrange the equation to solve for n: n = PV / RT Since the pressure remains constant, the change in the number of moles is zero.

We hope this solutions and analysis of the 2004 AP Chemistry FRQ Form B helps you prepare for your exam. Remember to practice with sample questions and review the concepts to achieve a high score.

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