Unit Study Guide: Solutions
A solution is a homogeneous mixture of two or more components. Homogeneous means the composition and properties are uniform throughout the entire mixture.
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Solvent: The component present in the largest quantity. It determines the physical state of the solution.
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Solute: The component(s) present in smaller quantities, dissolved in the solvent.
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Binary Solution: A solution containing exactly two components (one solute and one solvent).
1.1 Types of Solutions
Solutions aren't just liquids! Depending on the physical states of the solute and solvent, there are 9 different types of solutions, categorized into 3 major families:
| Solution Type | Solute (Minor) | Solvent (Major) | Common Real-World Example |
| Gaseous |
Gas Liquid Solid |
Gas Gas Gas |
Air (Mixture of $O_2$ and $N_2$) Chloroform mixed with nitrogen gas Camphor in nitrogen gas |
| Liquid |
Gas Liquid Solid |
Liquid Liquid Liquid |
Oxygen dissolved in water (Soda/Aquatic life) Ethanol (alcohol) dissolved in water Glucose or salt dissolved in water |
| Solid |
Gas Liquid Solid |
Solid Solid Solid |
Hydrogen gas absorbed in palladium Amalgam of mercury (liquid) with sodium Alloys like Copper dissolved in gold, or Brass |
1.2 Expressing Concentration of Solutions
Concentration tells us exactly how much solute is packed into a given amount of solution or solvent. Instead of vague words like "dilute" or "concentrated," we use quantitative units.
1. Mass Percentage ($\% \ w/w$)
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What it means: The mass of solute present in 100 grams of total solution.
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Formula:
$\text{Mass } \% \text{ of a component}$ $= \frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}} \times 100$
Easy Mental Picture: If a bottle says 10% Glucose in water by mass, it means you take 10 g of Glucose and mix it with 90 g of water to make a perfect 100 g solution.
2. Volume Percentage ($\% \ V/V$)
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What it means: The volume of liquid solute present in 100 mL of total solution.
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Formula:
$\text{Volume } \% \text{ of a component}$ $= \frac{\text{Volume of the component}}{\text{Total volume of solution}} \times 100$
Easy Mental Picture: A 35% (v/v) solution of ethylene glycol (car antifreeze) means 35 mL of antifreeze is diluted with water until the total volume reaches 100 mL.
3. Mass by Volume Percentage ($\% \ w/V$)
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What it means: The mass of solute (in grams) dissolved in 100 mL of the final solution.
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Where it's used: Very common in medicine, pharmacy, and IV fluid labels.
4. Parts Per Million (ppm)
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What it means: Used when a solute is present in microscopic trace amounts (like pollution in air or chlorine in a pool). It counts how many parts of solute exist per one million parts of solution.
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Formula:
$\text{Parts per million (ppm)}$ $= \frac{\text{Number of parts of the component}}{\text{Total number of parts of all components}} \times 10^6$
5. Mole Fraction ($x$)
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What it means: The ratio of the number of moles of one component to the total number of moles in the solution. It has no units.
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Formula for Binary Solution (Components A and B):
$$x_A = \frac{n_A}{n_A + n_B}$$$$x_B = \frac{n_B}{n_A + n_B}$$(Where $n_A$ and $n_B$ are the number of moles of A and B).
Golden Rule of Mole Fraction: The sum of all mole fractions always equals 1!
$$x_A + x_B = 1$$
Master the Numerical Problems (Step-by-Step)
Let's break down how to approach these calculations without getting overwhelmed.
Problem 1: Mass Percentage
Question: A solution is prepared by dissolving 15 g of sodium chloride (NaCl) in 135 g of water. Calculate the mass percentage of the solute.
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Step 1: Identify what is given.
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Mass of solute (NaCl) = 15 g
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Mass of solvent ($\text{H}_2\text{O}$) = 135 g
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Step 2: Calculate total mass of solution.
$\text{Total Mass} = \text{Solute} + \text{Solvent}$ $= 15\text{ g} + 135\text{ g} = 150\text{ g}$ -
Step 3: Plug into the formula.
$\text{Mass } \%$ $= \frac{15}{150} \times 100$ $= 0.1 \times 100 = 10\%$ -
Answer: The solution is 10% NaCl by mass.
Problem 2: Mole Fraction
Question: Calculate the mole fraction of ethylene glycol ($\text{C}_2\text{H}_6\text{O}_2$) in a solution containing 20% of $\text{C}_2\text{H}_6\text{O}_2$ by mass.
(Molar masses: $\text{C}_2\text{H}_6\text{O}_2 = 62\text{ g/mol}$, $\text{H}_2\text{O} = 18\text{ g/mol}$)
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Step 1: Simplify the percentage assumption.
Assume you have exactly 100 g of total solution.
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Mass of ethylene glycol = 20 g
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Mass of water = $100\text{ g} - 20\text{ g} =$ 80 g
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Step 2: Convert masses to moles ($n = \frac{\text{mass}}{\text{molar mass}}$).
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Moles of glycol ($n_{\text{glycol}}$) = $\frac{20\text{ g}}{62\text{ g/mol}} = 0.322\text{ moles}$
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Moles of water ($n_{\text{water}}$) = $\frac{80\text{ g}}{18\text{ g/mol}} = 4.444\text{ moles}$
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Step 3: Calculate Total Moles.
$$\text{Total Moles} = 0.322 + 4.444 = 4.766\text{ moles}$$ -
Step 4: Find Mole Fraction ($x$).
$$x_{\text{glycol}} = \frac{0.322}{4.766} = 0.068$$ -
Pro-Tip Bonus: If you want the mole fraction of water, you don't need to do the heavy division again! Just subtract from 1:
$$x_{\text{water}} = 1 - 0.068 = 0.932$$
Problem 1.1: Mass Percentage
Question: Calculate the mass percentage of benzene ($\text{C}_6\text{H}_6$) and carbon tetrachloride ($\text{CCl}_4$) if $22\text{ g}$ of benzene is dissolved in $122\text{ g}$ of carbon tetrachloride.
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The Goal: Find out what fraction of the total weight belongs to each chemical, written as a percentage.
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Step 1: Calculate the total mass of the solution.
$\text{Total Mass}$ $= \text{Mass of Benzene} + \text{Mass of CCl}_4$$\text{Total Mass}$ $= 22\text{ g} + 122\text{ g} = 144\text{ g}$ -
Step 2: Calculate the Mass % of Benzene.
$\text{Mass } \% \text{ of Benzene}$
$= \frac{22\text{ g}}{144\text{ g}} \times 100 = 15.28\%$ -
Step 3: Calculate the Mass % of $\text{CCl}_4$.
(You can either use the formula or simply subtract from 100%)
$\text{Mass } \% \text{ of CCl}_4$ $= 100\% - 15.28\% = 84.72\%$
Problem 1.2: Mole Fraction
Question: Calculate the mole fraction of benzene in a solution containing $30\%$ by mass in carbon tetrachloride.
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The Concept: A "30% by mass" solution means if you have a $100\text{ g}$ sample, $30\text{ g}$ is benzene and the remaining $70\text{ g}$ is carbon tetrachloride ($\text{CCl}_4$).
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Step 1: Determine the Molar Masses.
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Molar mass of Benzene ($\text{C}_6\text{H}_6$) = $(6 \times 12) + (6 \times 1) = 78\text{ g/mol}$
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Molar mass of $\text{CCl}_4$ = $12 + (4 \times 35.5) = 154\text{ g/mol}$
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Step 2: Convert masses to Moles ($n = \frac{\text{Mass}}{\text{Molar Mass}}$).
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$\text{Moles of Benzene } (n_{\text{benzene}})$ $= \frac{30\text{ g}}{78\text{ g/mol}}$ $= 0.385\text{ mol}$
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$\text{Moles of CCl}_4 \ (n_{\text{ccl4}})$ $= \frac{70\text{ g}}{154\text{ g/mol}}$ $= 0.455\text{ mol}$
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Step 3: Calculate Mole Fraction ($x$).
$x_{\text{benzene}}$ $= \frac{n_{\text{benzene}}}{n_{\text{benzene}} + n_{\text{ccl4}}}$ $= \frac{0.385}{0.385 + 0.455}$ $= \frac{0.385}{0.840} = 0.458$
Problem 1.3: Molarity Calculations
Question: Calculate the molarity of each of the following solutions:
(a) $30\text{ g}$ of $\text{Co(NO}_3)_2 \cdot 6\text{H}_2\text{O}$ in $4.3\text{ L}$ of solution.
(b) $30\text{ mL}$ of $0.5\text{ M H}_2\text{SO}_4$ diluted to $500\text{ mL}$.
Solution for (a):
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Step 1: Find the molar mass of the complete hydrated compound $\text{Co(NO}_3)_2 \cdot 6\text{H}_2\text{O}$.
$\text{Molar Mass}$ $= 58.9\text{ (Co)} + 2 \times [14\text{ (N)} + (3 \times 16\text{ (O)})] + 6 \times [18\text{ (H}_2\text{O)}]$$\text{Molar Mass}$ $= 58.9 + 124 + 108$ $= 290.9\text{ g/mol}$ -
Step 2: Find moles of solute.
$\text{Moles}$ $= \frac{30\text{ g}}{290.9\text{ g/mol}}$ $= 0.103\text{ mol}$ -
Step 3: Calculate Molarity ($\text{M} = \frac{\text{moles}}{\text{Volume in L}}$).
$\text{Molarity}$ $= \frac{0.103\text{ mol}}{4.3\text{ L}}$ $= 0.024\text{ M}$
Solution for (b): Dilution Formula
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The Rule: When you add water to dilute a solution, the total moles of solute stay the same ($M_1V_1 = M_2V_2$).
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Initial Concentration ($M_1$) = $0.5\text{ M}$, Initial Volume ($V_1$) = $30\text{ mL}$
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Final Volume ($V_2$) = $500\text{ mL}$, Final Concentration ($M_2$) = ?
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Calculation:
$$M_1V_1 = M_2V_2 \implies 0.5 \times 30 = M_2 \times 500$$$$15 = M_2 \times 500 \implies M_2 = \frac{15}{500} = 0.03\text{ M}$$
Problem 1.4: Working Backwards with Molality
Question: Calculate the mass of urea ($\text{NH}_2\text{CONH}_2$) required in making $2.5\text{ kg}$ of $0.25\text{ molal}$ aqueous solution.
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Understanding the Definition: A $0.25\text{ molal}$ solution means there are $0.25\text{ moles}$ of urea dissolved in exactly $1\text{ kg}$ ($1000\text{ g}$) of water.
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Step 1: Find mass of urea in this standard definition.
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Molar mass of urea ($\text{NH}_2\text{CONH}_2$) = $14 + 2 + 12 + 16 + 14 + 2 = 60\text{ g/mol}$
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$\text{Mass of 0.25 moles of urea} = 0.25 \times 60 = 15\text{ g}$
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Step 2: Find the total mass of this standard solution.
$\text{Total Mass}$ $= \text{Solute} + \text{Solvent}$ $= 15\text{ g (urea)} + 1000\text{ g (water)}$ $= 1015\text{ g} = 1.015\text{ kg}$ -
Step 3: Scale up to find what's needed for a $2.5\text{ kg}$ solution.
Using a simple proportion:
$\text{Mass of Urea required}$ $= \frac{15\text{ g of urea}}{1.015\text{ kg of solution}} \times 2.5\text{ kg of solution}$ $= 36.95\text{ g}$
Problem 1.5: The Triple Challenge (Molality, Molarity, & Mole Fraction)
Question: Calculate (a) molality, (b) molarity, and (c) mole fraction of $\text{KI}$ if the density of $20\%$ (mass/mass) aqueous $\text{KI}$ is $1.202\text{ g mL}^{-1}$.
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Base Assumption: Assume you have a $100\text{ g}$ total solution.
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Mass of solute ($\text{KI}$) = $20\text{ g}$
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Mass of solvent ($\text{H}_2\text{O}$) = $100\text{ g} - 20\text{ g} = 80\text{ g} = 0.08\text{ kg}$
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Molar mass of $\text{KI}$ = $39 + 127 = 166\text{ g/mol}$
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$\text{Moles of KI} = \frac{20\text{ g}}{166\text{ g/mol}} = 0.1205\text{ mol}$
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(a) Calculate Molality ($m$):
(b) Calculate Molarity ($M$):
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First, find the volume of the $100\text{ g}$ solution using density ($\text{Volume} = \frac{\text{Mass}}{\text{Density}}$):
$$\text{Volume} = \frac{100\text{ g}}{1.202\text{ g/mL}} = 83.19\text{ mL} = 0.08319\text{ L}$$ -
Now, compute molarity:
$M = \frac{\text{Moles of Solute}}{\text{Volume of Solution in L}}$ $= \frac{0.1205\text{ mol}}{0.08319\text{ L}}$ $= 1.448\text{ M}$
(c) Calculate Mole Fraction ($x_{\text{KI}}$):
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Find moles of water: $\text{Moles of water} = \frac{80\text{ g}}{18\text{ g/mol}} = 4.444\text{ mol}$
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Compute mole fraction:
$x_{\text{KI}}$ $= \frac{\text{Moles of KI}}{\text{Moles of KI} + \text{Moles of Water}}$ $= \frac{0.1205}{0.1205 + 4.444}$ $= 0.026$
Section 1.3.1: Solubility of a Solid in a Liquid
1.3.1 Solubility of a Solid in a Liquid
Solubility is defined as the maximum amount of a solute that can be dissolved in a specified amount of solvent at a specific temperature.
When a solid dissolves in a liquid, it doesn't happen randomly. It is governed by chemical rules, equilibrium, temperature, and pressure.
1. The Core Rule: "Like Dissolves Like"
Every solid does not dissolve in every liquid. Dissolution depends heavily on the nature of the solute and the solvent.
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Polar Solutes dissolve in Polar Solvents: Substances with charges or poles (like Sodium Chloride/Salt, Sugar) dissolve easily in polar liquids like Water.
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Non-polar Solutes dissolve in Non-polar Solvents: Substances without charges (like Naphthalene or Anthracene) will not dissolve in water, but they readily dissolve in non-polar liquids like Benzene.
Summary: Intermolecular interactions must be similar between the solute and solvent for dissolving to happen.
2. The Dynamic Equilibrium (How Dissolving Works)
When you drop a solid into a liquid, two competing processes begin:
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Dissolution: Solute particles break away from the solid and mix into the solvent. The concentration of solute in the solution increases.
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Crystallisation: Dissolved solute particles bump into each other or the remaining solid, crashing out of the solution to become solid again.
Eventually, a state of Dynamic Equilibrium is reached where the speed of both processes becomes exactly equal:
Saturated vs. Unsaturated Solutions
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Saturated Solution: A solution that has reached this equilibrium point. It contains the maximum possible amount of dissolved solute at that specific temperature and pressure. No more solute can be dissolved.
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Unsaturated Solution: A solution in which more solute can still be dissolved at that temperature.
3. Factors Affecting Solid-in-Liquid Solubility
A. Temperature (The Deciding Factor)
Because a saturated solution exists in a state of dynamic equilibrium, it follows Le Chatelier’s Principle. The effect of temperature depends entirely on whether dissolving heat is absorbed or released:
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Endothermic Dissolution ($\Delta_{\text{sol}}H > 0$): If the process absorbs heat (needs warmth to dissolve), increasing the temperature increases solubility.
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Exothermic Dissolution ($\Delta_{\text{sol}}H < 0$): If the process releases heat, increasing the temperature decreases solubility.
B. Pressure (The No-Effect Factor)
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Effect: Practically none.
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Why? Solids and liquids are highly incompressible. Changing the external pressure does not change their volume or force more solid particles into the liquid structure.
Quick-Review Summary Table
| Factor | Condition / Type | Effect on Solubility of Solid in Liquid |
| Nature of Substance | Similar polarities ("Like dissolves Like") | High Solubility |
| Nature of Substance | Mismatched polarities (Polar + Non-polar) | Insoluble / Low Solubility |
| Temperature | Endothermic ($\Delta H = +ve$) | Increases with higher temperature |
| Temperature | Exothermic ($\Delta H = -ve$) | Decreases with higher temperature |
| Pressure | Any | No significant effect |