Lesson 1 AP CHEMISTRY SABIS

​SABIS AP Chemistry Lesson 1

By Dr. Hisham Mahmoud

• “Welcome to AP Chemistry 2025! I’m Mr. Hisham. My goal: make every calculation crystal clear.”

By Dr. Hisham Mahmoud

• Welcome to The Last Boss Level

By Dr. Hisham Mahmoud

• When you pass this year , College will be Candy Crush for you🍬🍫

By Dr. Hisham Mahmoud

• We’ll always start small → build step by step → reach exam-level problems.

By Dr. Hisham Mahmoud

We will Review Chapters 1-5

As Always Our Source is school material , that is from Zumdahl Chemistry Book For school

Let's Try to know you
a little bit everytime

Word Cloud

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Now, let’s see if you can answer THIS one…

25 cm³ of a salt solution leaves just 1.11 g solid when evaporated. How can this tiny leftover tell us the identity of the salt?

25 cm³ of a salt solution leaves just 1.11 g solid when evaporated. How can this tiny leftover tell us the identity of the salt?

If you know 3, you can always find the 4th.

Multiple Choice

25.0 cm³ of 0.400 M salt solution evaporates → 1.11 g solid left. Find molar mass..

Which quantity is the residue mass

• • 0.400 g

• • 25.0 g

• • 1.11 g

• • 0.0250 g

• Residue means what’s left — solid or solution?

Multiple Choice

part 2

To connect mass (m) and molar mass (M), which equation is correct?

• • m = n × M

• • m = n ÷ M

• • m = M ÷ n

• • M = 1 ÷ CV

​What is the unit of molar mass?” →

Multiple Choice

Part 3 Moles in Solution
if m = n X M

Which expression gives the moles of solute in this solution?

• • n = C × V

• • n = m ÷ M

• • n = M × V

• • n = C ÷ V

Multiple Choice

Part 4
Combining Equations

If m = C × V × M, how do we rearrange for M

• • M = $$\frac{CV}{M}$$

• • M = $$\frac{m}{CV}$$

• • M = $$m\times C\times V$$

• • M = $$\frac{C}{mV}$$

Lost? Let’s recap 👇

How Did We Reach M

Multiple Choice

Lost? Let’s recap

• We already know:

  • m (mass) = 1.11 g

  • C (concentration) = 0.400 M

  • V (volume) = 25.0 cm³ = 0.0250 L

• Formula we need: M = m ÷ (C × V)

• • M = $$\frac{CV}{M}$$

• • M = $$\frac{m}{CV}$$

• • M = $$m\times C\times V$$

• • M = $$\frac{C}{mV}$$

Multiple Choice

Quick Check Q:

Which step introduced the connection between solution concentration and moles?

• • Step 1

• • Step 2

• • Step 3

• • Step 4

Multiple Choice

What is the formula for moles from concentration and volume?

• • n = m ÷ M

• • n = C × V

• • n = V ÷ C

• • n = C ÷ m

Word Cloud

• So what do we call 111 here — mass or molar mass?

Multiple Choice

M = ?

• • 44.4

• • 222 g/mol

• • 111 g/mol

• • 400 g/mol

Remember, multiply C×V first: 0.400 x 0.0250 = 0.0100.

Then 1.11 ÷ 0.0100 = 111.

Multiple Choice

Why did we convert 25.0 cm³ → 0.0250 L?

• • To change density

• • To match grams

• • To avoid decimals

• • Molarity uses liters

Multiple Choice

Given 0.400, 25.0, 1.11 (all 3 s.f.), report M as

• • 111.0 g/mol

• • 111 g/mol

• • 1.11×10² g

• • 1.1×10² g/mol

Multiple Choice

You found M ≈ 111 g/mol. Which formula best fits?

• • NaCl (58.44)

• • KCl (74.55)

• • CaCl₂ (110.98)

• • Na₂CO₃ (106.0)

​🧪 Problem:

2.12 g of Na₂CO₃ dissolved in 500 cm³ solution. Calculate concentrations of all ions.

Multiple Choice

Step 1 (focus check): Which formula calculates moles from mass?

• • n = C × V

• • n = m ÷ M

• • m = n × M

• • M = m ÷ V

Fill in the Blanks

Fill in the Blanks

Fill in the Blanks

​Why must we convert 500 cm³ to 0.500 L ?

• Because molarity uses liters ✅

Fill in the Blanks

​We now know:
[Na₂CO₃] = 0.0400 M.

Next, split into ions!

Multiple Choice

Step 5

• Na₂CO₃ (aq) → 2 Na⁺ (aq) + CO₃²⁻ (aq)

How many sodium ions form per formula unit?

• 1

• 2

• 3

• 4

Multiple Choice

Step 6

• Na₂CO₃ (aq) → 2 Na⁺ (aq) + CO₃²⁻ (aq)

if [Na₂CO₃] = 0.0400

what is

[Na⁺] and [CO₃²⁻] in M ?

• • 0.0800

  and

  • 0.0800

• • 0.0800

  and

  • 0.0400

• • 0.0400

  and

  • 0.0400

• • 0.0400

  and

  • 0.0800

Multiple Choice

Step 1 — Identify moles of NaCl

Which formula gives moles from concentration and volume ?v

• • n = C × V

• • n = m ÷ M

Fill in the Blanks

Fill in the Blanks

• Remember: in CaCl₂
  2 Cl^- ions per 1 formula unit →double the concentration

Multiple Choice

Step 4 — How many moles of Cl⁻ come from CaCl₂?
remember ratio is 1:2

• 1.5 x 10^-3

• 3.0 x 10^-3

Multiple Choice

Step 5 now we need to know the total Cl⁻ ions from

• NaCl → Na⁺ + Cl⁻

• CaCl₂ → Ca²⁺ + 2Cl⁻
  what to do ?

• add [Cl^-] from both
  
  (5.0 × 10-3) + 2(1.5 × 10-3 mole) =

• multiply [Cl^-] from both
  
  (5.0 × 10-3) x 2(1.5 × 10-3 mole) =

Multiple Choice

Step 6

Answer total

Cl⁻

is .... x10^-3.

• 1.0 x 10^-3

• 3.0 x 10^-3

• 8.0 x 10^-3

• 2.0 x 10^-3

Remember this is the no of moles of chloride ions
not its concentration

To find concentration divide the total no of chloride moles by the total volume of the solution after mixing .

Multiple Choice

step 7

After mixing 25 cm³ + 75 cm³, the total volume in liters is:

• • 0.0100 L

• • 0.100 L

• • 1.00 L

• • 0.075 L

Multiple Choice

step 8

Cl^- = (8.0 x 10⁻³) / 0.100

• • 8.0 x 10⁻²

• • 4.0 x 10⁻²

• • 2.0 x 10⁻²

• • 1.0 x 10⁻²

🧪

Problem:

A lab technician is asked to prepare 500 cm³ of 0.50 M H₂SO₄ from a 6.0 M stock solution.

Multiple Choice

👉 Which formula links concentration and volume for dilutions?

• • C₁V₁ = C₂V₂

• • m = n × M

• • n = m ÷ M

• • pV = nRT

Multiple Choice

Convert Target Volume

Target solution = 500 cm³ = 0.500 dm³

Why convert to dm³

• • molarity uses liters

• • molarity uses cm³

Multiple Choice

How many moles of H₂SO₄ are required?

n = C × V = (0.50)(0.500) = 0.25 mol

• • 0.25 mol

• • 0.025 mol

Multiple Choice

Volume of Stock

?

n ÷ C = 0.25 ÷ 6.0 = 0.0417 dm³

• • 4.17 cm³

• • 41.7 cm³

• • 417 cm³

• • 0.417 cm³

Multiple Choice

Using a burette, measure 41.7 cm³ of 6.0 M H₂SO₄.

👉 Focus Q: Why use a burette instead of a beaker?

• • Safer

• • Faster

• • Larger volume

• • More accurate

Place ~50 ml of distilled water in a 500 cm³ volumetric flask. Add the measured acid slowly.

Acid to water or water to acid first?

• Acid into water ✅ (safety rule).

Thank You 😊