General Chemistry Chapter 1 Practice Problems

General Chemistry Chapter 1 Practice Problems: A Comprehensive Guide

General chemistry, often the first hurdle in a science student's journey, introduces fundamental concepts crucial for understanding more advanced chemistry topics. Chapter 1 typically lays the groundwork, covering essential mathematical tools, units, significant figures, and basic problem-solving strategies. Mastering these foundational skills is paramount for success in the rest of the course. This article provides a comprehensive guide to common Chapter 1 practice problems, offering detailed solutions and explanations to build your confidence and understanding.

Section 1: Measurement and Significant Figures

This section focuses on problems involving units, unit conversions, and significant figures – the language of chemistry. Accuracy and precision are key.

1.1 Unit Conversions:

Problem: Convert 2500 centimeters (cm) to kilometers (km).

Solution: We use the following conversion factors: 1 m = 100 cm and 1 km = 1000 m.

1. Start with the given value: 2500 cm
2. Apply conversion factors: 2500 cm * (1 m / 100 cm) * (1 km / 1000 m)
3. Cancel units: The 'cm' and 'm' units cancel out, leaving km.
4. Calculate: 2500 / (100 * 1000) = 0.025 km

Answer: 2500 cm is equal to 0.025 km.

1.2 Significant Figures:

Problem: Perform the following calculation and report the answer with the correct number of significant figures: (2.55 x 10³ g) + (1.2 x 10² g)

Solution:

1. Convert to the same power of 10: 2550 g + 120 g
2. Perform the addition: 2670 g
3. Determine significant figures: 1.2 x 10² g has two significant figures, which limits the precision of the sum. The least precise measurement dictates the number of significant figures in the final answer.
4. Express the answer with correct significant figures: 2.7 x 10³ g (two significant figures).

Answer: The correct answer is 2.7 x 10³ g.

1.3 Scientific Notation and Precision:

Problem: Express 0.0000456 in scientific notation and state the number of significant figures.

Solution:

1. Move the decimal point to obtain a number between 1 and 10: 4.56
2. Count the number of places the decimal point was moved: The decimal point was moved five places to the right.
3. Write the number in scientific notation: 4.56 x 10⁻⁵
4. Determine the number of significant figures: The number has three significant figures (4, 5, and 6).

Answer: 0.0000456 is expressed as 4.56 x 10⁻⁵ with three significant figures.

Section 2: Dimensional Analysis and Problem Solving

Dimensional analysis, also known as the factor-label method, is a powerful technique to solve complex chemistry problems.

2.1 Density Calculations:

Problem: A sample of gold has a mass of 193.2 g and a volume of 10.0 cm³. Calculate the density of gold.

Solution: Density is defined as mass per unit volume: Density = Mass / Volume

1. Identify the given values: Mass = 193.2 g, Volume = 10.0 cm³
2. Apply the formula: Density = 193.2 g / 10.0 cm³
3. Calculate: Density = 19.32 g/cm³
4. Significant Figures: The volume (10.0 cm³) has three significant figures, so the density should also have three significant figures.

Answer: The density of gold is 19.3 g/cm³.

2.2 Using Multiple Conversion Factors:

Problem: Convert 60 miles per hour (mph) to meters per second (m/s). Use the following conversion factors: 1 mile = 1609 meters, 1 hour = 60 minutes, 1 minute = 60 seconds.

Solution:

1. Start with the given value: 60 mph
2. Apply conversion factors sequentially: 60 mph * (1609 m / 1 mile) * (1 hour / 60 min) * (1 min / 60 s)
3. Cancel units: The units of miles, hours, and minutes cancel out, leaving m/s.
4. Calculate: (60 * 1609) / (60 * 60) = 26.8 m/s

Answer: 60 mph is approximately equal to 26.8 m/s.

Section 3: Introduction to Matter and Its Properties

This section introduces basic concepts of matter, including states of matter, physical and chemical properties, and changes.

3.1 Classifying Matter:

Problem: Classify the following as a pure substance, homogeneous mixture, or heterogeneous mixture: (a) air, (b) distilled water, (c) granite.

Solution:

• (a) Air: Air is a homogeneous mixture of gases (primarily nitrogen, oxygen, and argon).
• (b) Distilled water: Distilled water is a pure substance (specifically, a compound).
• (c) Granite: Granite is a heterogeneous mixture of minerals (quartz, feldspar, mica).

Answer: (a) Homogeneous mixture, (b) Pure substance, (c) Heterogeneous mixture.

3.2 Physical vs. Chemical Changes:

Problem: Identify each of the following as a physical or chemical change: (a) boiling water, (b) burning wood, (c) dissolving sugar in water.

Solution:

• (a) Boiling water: A physical change. The water changes state from liquid to gas, but its chemical composition remains the same (H₂O).
• (b) Burning wood: A chemical change. The wood reacts with oxygen in the air, producing new substances (carbon dioxide, water vapor, ash).
• (c) Dissolving sugar in water: A physical change. The sugar dissolves and forms a solution but doesn't undergo a chemical reaction; it can be recovered unchanged.

Answer: (a) Physical change, (b) Chemical change, (c) Physical change.

Section 4: Atomic Structure and the Periodic Table

4.1 Atomic Number and Mass Number:

Problem: An atom of oxygen has 8 protons and 8 neutrons. What are its atomic number and mass number?

Solution:

• Atomic number: The atomic number is equal to the number of protons, which is 8.
• Mass number: The mass number is the sum of protons and neutrons, which is 8 + 8 = 16.

Answer: Atomic number = 8, Mass number = 16.

4.2 Isotopes and Atomic Mass:

Problem: Chlorine has two isotopes: ³⁵Cl (75.77% abundance) and ³⁷Cl (24.23% abundance). Calculate the average atomic mass of chlorine.

Solution:

1. Convert percentages to decimal form: 75.77% = 0.7577, 24.23% = 0.2423
2. Multiply each isotope's mass by its abundance: (35 amu * 0.7577) + (37 amu * 0.2423)
3. Add the results: 26.5195 amu + 8.9651 amu = 35.4846 amu
4. Round to the correct number of significant figures: 35.48 amu

Answer: The average atomic mass of chlorine is 35.48 amu.

Section 5: Introduction to the Mole Concept

The mole is a crucial concept in chemistry that bridges the microscopic world of atoms and molecules to the macroscopic world of measurable quantities.

5.1 Mole-to-Gram Conversions:

Problem: Calculate the mass in grams of 2.50 moles of water (H₂O). The molar mass of water is 18.02 g/mol.

Solution:

1. Identify the given values: Moles = 2.50 mol, Molar mass = 18.02 g/mol
2. Use the formula: Mass (g) = Moles * Molar mass
3. Calculate: Mass = 2.50 mol * 18.02 g/mol = 45.05 g

Answer: The mass of 2.50 moles of water is 45.05 g.

5.2 Gram-to-Mole Conversions:

Problem: How many moles are present in 50.0 g of sodium chloride (NaCl)? The molar mass of NaCl is 58.44 g/mol.

Solution:

1. Identify the given values: Mass = 50.0 g, Molar mass = 58.44 g/mol
2. Use the formula: Moles = Mass (g) / Molar mass
3. Calculate: Moles = 50.0 g / 58.44 g/mol = 0.855 mol

Answer: There are 0.855 moles in 50.0 g of NaCl.

Conclusion:

This comprehensive guide covers a wide range of practice problems typically encountered in General Chemistry Chapter 1. Remember, consistent practice is key to mastering these fundamental concepts. Work through numerous problems, focusing on understanding the underlying principles and applying the appropriate techniques. Don't hesitate to seek help from your instructor, textbooks, or online resources if you encounter difficulties. By diligently practicing, you will build a solid foundation for success in your chemistry studies. Good luck!