Chemistry288Updated 26 Oct 202612 pages

Understanding Moles in Solution - AQA Chemistry Topic 2

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Getting your head around moles in solution is crucial for chemistry calculations - it's basically about figuring out how much stuff is dissolved in a given amount of liquid. This topic connects the mole concept you've learned with real-world solutions, giving you the tools to calculate concentrations and work backwards from them too.
2.2 Moles in solution – page 1

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Understanding Solutions and Concentration

Think of making a cup of tea - you've got water (the solvent) and sugar (the solute), and when mixed together you get your solution. That's exactly what chemists work with, just with different substances.

Concentration tells us how much solute is packed into a specific volume of solution - not just the solvent alone. It's like asking "how strong is this solution?" The higher the concentration, the more particles are crammed into each litre.

Quick Tip: Remember that concentration is about the final solution volume, not just the amount of water you started with!

2.2 Moles in solution – page 2

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The Concentration Formula

The key formula you need to master is: concentration (mol dm⁻³) = moles ÷ volume (dm³). This simple equation is your best friend for solution calculations.

A molar solution (1M) contains exactly 1 mole of substance dissolved in 1 dm³ of solution. So a 3M solution has 3 moles per dm³ - three times more concentrated. The units mol dm⁻³ might look intimidating, but they just mean "moles per cubic decimetre".

Here's what's crucial: 1 mol dm⁻³ literally means 1 mole of solute in every cubic decimetre of solution. Once you get this concept, the calculations become straightforward.

Key Conversion: Always remember that 1 dm³ = 1000 cm³ - you'll use this constantly!

2.2 Moles in solution – page 3

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Worked Example and Formula Rearrangements

Let's tackle a proper example: 1.17g of sodium chloride dissolved in 500 cm³ of solution. First, find the moles: 1.17g ÷ 58.5 g/mol = 0.020 mol. Then convert volume to dm³: 500 cm³ = 0.500 dm³.

Using our formula: concentration = 0.020 ÷ 0.500 = 0.040 mol dm⁻³. Simple when you break it down step by step!

The formula can be rearranged depending on what you're looking for. If you need moles from a solution, use: moles = concentration × volume (dm³). For cm³ volumes, use: moles = (concentration × volume in cm³) ÷ 1000.

Pro Tip: Always convert cm³ to dm³ by dividing by 1000 - it's the most common mistake to avoid!

2.2 Moles in solution – page 4

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Key Definitions and Conversions

Molarity is just another word for molar concentration - it's the moles of solute per litre of solution. The formula stays the same: M = n ÷ v, where M is molarity, n is moles, and v is volume in litres.

The volume conversion is absolutely essential: 1 dm³ = 1000 cm³. To convert from dm³ to cm³, multiply by 1000. To go the other way, divide by 1000.

This conversion trips up loads of students, so practise it until it's automatic. Most exam questions give volumes in cm³, but the concentration formula works best with dm³.

Memory Hook: Think "dm³ is bigger, so the number is smaller" - 0.5 dm³ = 500 cm³.

2.2 Moles in solution – page 5

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Formula Triangle and Layout

The concentration relationship can be visualised with the classic triangle method. Put concentration at the top, then moles and volume at the bottom. Cover what you want to find, and you'll see how to calculate it.

The standard formula layout is: C = n ÷ V, where C is concentration in mol/dm³, n is amount in moles, and V is volume in dm³. This gives you concentration directly.

Whether you're finding concentration, moles, or volume, this same relationship applies. Just rearrange algebraically - cover the unknown, and the triangle shows you the calculation.

Visual Learner Tip: Draw the triangle in your exam - it takes 5 seconds and prevents formula mix-ups!

2.2 Moles in solution – page 6

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Practice Calculations

Here are the calculation patterns you'll see repeatedly. For finding moles: moles = (1.2 × 25) ÷ 1000 = 0.030 moles from a 1.2 mol dm⁻³ solution.

For finding concentration: 0.2 moles in 0.4 dm³ gives 0.2 ÷ 0.4 = 0.5 mol dm⁻³. For finding volume: 0.080 moles at 0.50 mol dm⁻³ means 0.080 ÷ 0.50 = 0.16 dm³.

When questions ask for mass, work backwards: find moles first (2.00 mol dm⁻³ × 0.050 dm³ = 0.100 mol), then multiply by Mr 0.100×40.0=4.00g0.100 × 40.0 = 4.00g.

Exam Strategy: Always write out what you're given and what you need to find - it prevents silly mistakes under pressure!

2.2 Moles in solution – page 7

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Step-by-Step Method

The step-by-step approach helps you understand where the formula comes from. If 1000 cm³ of 1.00 mol dm⁻³ contains 1 mol, then 1000 cm³ of 0.100 mol dm⁻³ contains 0.100 mol.

Therefore, 1.0 cm³ contains 0.100 ÷ 1000 = 0.0001 mol. So 25.0 cm³ contains 25.0 × 0.0001 = 0.0025 mol. This matches the formula: (0.10 × 25.0) ÷ 1000 = 0.0025 mol.

For the KCl example: molar mass = 39.1 + 35.5 = 74.6 g/mol. Then 111.9g ÷ 74.6 g/mol = 1.5 moles.

Understanding Check: If you can explain why we divide by 1000, you've properly understood the concept!

2.2 Moles in solution – page 8

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Completing Multi-Step Problems

The final step for our KCl problem: concentration = 1.5 moles ÷ 0.5 dm³ = 3 mol dm⁻³. Notice how we systematically work through: molar mass → moles → concentration.

For magnesium iodide (MgI₂): molar mass = 24.3 + (2 × 126.9) = 278.1 g/mol. This gives us the foundation for the moles calculation in the next step.

The pattern is always the same: find molar mass, calculate moles from the given mass, then find concentration using the volume. Master this sequence and you'll ace solution calculations.

Time-Saver: Write out the molar mass calculation clearly - it's easy marks and prevents arithmetic errors!

2.2 Moles in solution – page 9

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Final Worked Examples

Continuing the MgI₂ calculation: 55.62g ÷ 278.1 g/mol = 0.2 moles. Then concentration = 0.2 moles ÷ 0.4 dm³ = 0.5 mol dm⁻³.

For the sodium nitrate example, start with the molar mass: Na (23.0) + N (14.0) + 3O (3 × 16.0) = 85.0 g/mol. Remember that NO₃ contains three oxygen atoms.

The volume is given as 125 cm³, so you'll need to convert this to dm³ 125÷1000=0.125dm3125 ÷ 1000 = 0.125 dm³ before calculating the final concentration.

Final Tip: Double-check your molar mass calculations - count atoms carefully, especially in compounds like NO₃!

2.2 Moles in solution – page 10

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Bringing It All Together

The final sodium nitrate calculation: 21.25g ÷ 85.0 g/mol = 0.25 moles. With volume = 125 cm³ = 0.125 dm³, concentration = 0.25 ÷ 0.125 = 2.0 mol dm⁻³.

Notice the consistent method: molar mass → moles → concentration. Whether the volume is in dm³ or cm³, whether you're finding concentration or working backwards to find mass, the same logical steps apply.

These calculations might seem complex at first, but they're just the mole concept applied to solutions. Once you've practised the method, you'll find them quite straightforward.

Success Strategy: Practise converting between cm³ and dm³ until it's automatic - it's the foundation skill for all solution calculations!

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