📖 Lesson 7: Preparation of Solutions

👩🏽‍🏫 Introduction

Hi lab star! 🧪✨

Have you ever mixed sugar in water to make a drink? Congratulations—you’ve prepared a solution! In science, solution preparation is an essential skill. Whether you’re a pharmacist making medicine or a farmer mixing fertilizer, knowing how to prepare a solution accurately is a must.

In this lesson, we’ll learn what a solution is, the parts of a solution, how to calculate concentration, and how to prepare specific volumes of solutions in real-life and lab situations.

📘 Core Concepts

🔹 What Is a Solution?

A solution is a uniform mixture of two or more substances.

🔹 Parts of a Solution:

• Solute: the substance that is dissolved (e.g. salt, sugar)

• Solvent: the substance that does the dissolving (e.g. water)

🧠 Solution = Solute + Solvent

🔹 Concentration of a Solution

Concentration tells us how much solute is in a given amount of solution.

🧠 Common unit: Molarity (mol/L) or Moles per litre (mol/dm³)

Formula:
Molarity (M) = Moles of solute ÷ Volume of solution in litres

🔹 Standard Solutions

A standard solution is one whose concentration is known exactly.

How to prepare one:

1. Weigh the exact mass of solute needed.

2. Dissolve it in a small volume of distilled water.

3. Transfer into a volumetric flask.

4. Add more distilled water up to the mark (e.g., 1.0 L).

🔹 Dilution of Solutions

To make a weaker solution from a strong one, use:

🧠 C₁V₁ = C₂V₂

Where:

• C₁ = initial concentration

• V₁ = volume of stock (strong) solution

• C₂ = final (new) concentration

• V₂ = final (new) volume

👩🏽‍🔬 Real-Life Examples

🧩 Sample Problem Walkthrough

🧠 Problem 1:

How many grams of sodium chloride (NaCl) are needed to prepare 250 cm³ of a 0.2 mol/dm³ solution?
(Molar mass of NaCl = 58.5 g/mol)

✅ Step-by-Step Solution:

Step 1: Use
Moles = Molarity × Volume (in litres)
= 0.2 mol/dm³ × 0.250 dm³ = 0.05 mol

Step 2:
Mass = Moles × Molar Mass
= 0.05 mol × 58.5 g/mol = 2.93 g

✔️ Final Answer: 2.93 grams of NaCl

🧠 Problem 2 (Dilution):

How much of a 1.0 mol/dm³ HCl solution is needed to make 100 cm³ of a 0.25 mol/dm³ solution?

Use: C₁V₁ = C₂V₂

C₁ = 1.0 mol/dm³
C₂ = 0.25 mol/dm³
V₂ = 100 cm³
V₁ = ?

V₁ = (C₂ × V₂) ÷ C₁ = (0.25 × 100) ÷ 1.0 = 25 cm³

✔️ Final Answer: Use 25 cm³ of the strong solution and dilute with water to make 100 cm³

✍🏽 Practice Exercises

Exercise 1: Fill-in-the-Blanks

1. The ______ is the substance that is dissolved in a solution.

2. Molarity is measured in ______.

3. To dilute a solution, we use the formula ______ = ______.

Answers:

1. solute

2. mol/dm³

3. C₁V₁ = C₂V₂

Exercise 2: Multiple Choice

1. What equipment is used to prepare a precise volume of solution?
   A. Beaker
   B. Test tube
   C. Measuring cylinder
   D. Volumetric flask

✅ Answer: D. Volumetric flask

2. If you dissolve 0.1 mol of NaOH in 1 litre of water, the concentration is:
   A. 0.01 mol/dm³
   B. 0.1 mol/dm³
   C. 1.0 mol/dm³
   D. 10 mol/dm³

✅ Answer: B. 0.1 mol/dm³

Exercise 3: Calculation

Question:
Calculate the mass of potassium nitrate (KNO₃, molar mass = 101 g/mol) needed to make 0.5 L of a 0.1 mol/dm³ solution.

Solution:
Moles = M × V = 0.1 × 0.5 = 0.05 mol
Mass = 0.05 × 101 = 5.05 g

🌀 Recap

Let’s review:

• A solution is made by dissolving a solute in a solvent.

• Molarity measures how concentrated the solution is:

  Molarity = Moles ÷ Volume

• You can prepare a standard solution by dissolving the correct mass in a known volume.

• To dilute a solution, use:

  C₁V₁ = C₂V₂

💭 Reflection Prompt

You’re asked to prepare a 500 cm³ sugar solution for an experiment. How would you go about it step by step? What tools would you need?
Write your plan as if you’re giving lab instructions to a friend.