Unit 1: Solutions
Introduction
A solution is a homogeneous mixture of two or more substances. The substance present in the larger amount is called the solvent, and the one present in smaller amount is called the solute. Solutions are all around us — sea water, air, vinegar, and alloys are common examples.
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Types of Solutions
- Solutions can exist in all three states of matter. The nine possible types arise from combining gaseous, liquid, and solid solutes with gaseous, liquid, and solid solvents. For example:
- Gas in liquid: oxygen dissolved in water
- Solid in liquid: sugar dissolved in water (most common)
- Gas in gas: air (mixture of N2, O2, etc.)
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Expressing Concentration of Solutions
1. Molarity (M): Number of moles of solute per litre of solution.
M = moles of solute / volume of solution in litres
2. Molality (m): Number of moles of solute per kilogram of solvent.
m = moles of solute / mass of solvent in kg
Molality does NOT change with temperature (unlike molarity).
3. Mole Fraction (x): Ratio of moles of one component to the total moles in the solution.
xA = nA / (nA + nB)
4. Mass by Mass Percentage (w/w): (mass of solute / mass of solution) x 100
5. Parts per million (ppm): mass of solute / mass of solution x 106. Used for very dilute solutions (e.g., water pollutants).
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Solubility
Solubility of a substance is the maximum amount of solute that can be dissolved in a given amount of solvent at a specified temperature.
- Henry's Law: At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
p = KH x x (where KH is Henrys law constant, x is mole fraction of gas)
Higher KH means lower solubility. This explains why gases like CO2 stay dissolved in cold soda but escape when pressure is reduced.
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Vapour Pressure of Solutions
The vapour pressure of a pure solvent decreases when a non-volatile solute is added. This is called lowering of vapour pressure.
Raoults Law (for ideal solutions):
pA = xA x pA · (where pA · is the vapour pressure of pure solvent)
Relative lowering of vapour pressure:
(pA · - pA) / pA · = xB (mole fraction of solute)
- Solutions that obey Raoults law at all concentrations are ideal solutions (e.g., benzene + toluene). Those that deviate are non-ideal:
- Positive deviation: Interactions between unlike molecules weaker than like molecules (e.g., ethanol + water). Vapour pressure > Raoults law prediction.
- Negative deviation: Unlike interactions stronger than like molecules (e.g., acetone + chloroform).
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Colligative Properties
These properties depend only on the number of solute particles, not their nature.
1. Relative Lowering of Vapour Pressure:
(p · - p) / p · = nB / (nA + nB)
2. Elevation in Boiling Point (delta Tb):
delta Tb = Kb x m
Kb is the ebullioscopic constant (molal boiling point elevation constant).
3. Depression in Freezing Point (delta Tf):
delta Tf = Kf x m
Kf is the cryoscopic constant. This is used to determine molar mass of solutes.
4. Osmotic Pressure (pi):
pi = CRT = (n/V) x R x T
Osmosis is flow of solvent from lower concentration to higher concentration through a semipermeable membrane.
Van't Hoff Factor (i):
For electrolytes that dissociate or associate, modified colligative properties use:
delta Tb = i x Kb x m
i = observed colligative property / theoretical (for non-electrolyte)
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Worked Examples
Calculate the molarity of a solution made by dissolving 4 g of NaOH (molar mass = 40 g/mol) in 500 mL of solution.
Moles of NaOH = 4/40 = 0.1 mol
Volume in litres = 500/1000 = 0.5 L
M = 0.1/0.5 = 0.2 M
Calculate the mole fraction of ethanol in a solution containing 46 g of ethanol (M = 46 g/mol) and 180 g of water (M = 18 g/mol).
Moles of ethanol = 46/46 = 1 mol
Moles of water = 180/18 = 10 mol
xethanol = 1/(1+10) = 1/11 ≈ 0.0909
The vapour pressure of pure water at 25°C is 23.8 mmHg. What is the vapour pressure when 10 g of glucose (M = 180) is dissolved in 90 g of water (M = 18)?
Moles glucose = 10/180 ≈ 0.0556; Moles water = 90/18 = 5
xwater = 5/(5 + 0.0556) = 5/5.0556 ≈ 0.989
p = 0.989 x 23.8 ≈ 23.54 mmHg
Calculate the boiling point elevation when 1.5 g of urea (M = 60) is dissolved in 100 g of water. Kb for water = 0.52 K kg/mol.
Moles of urea = 1.5/60 = 0.025 mol
Molality m = 0.025/0.1 kg = 0.25 mol/kg
delta Tb = 0.52 x 0.25 = 0.13 K
Calculate the osmotic pressure of a 0.01 M glucose solution at 27°C. (R = 0.0821 L atm mol-1 K-1)
T = 27 + 273 = 300 K
pi = CRT = 0.01 x 0.0821 x 300 = 0.2463 atm
The freezing point of a solution containing 0.6 g of urea in 100 g of water is depressed by 0.186°C. Find the molar mass of urea. Kf for water = 1.86 K kg/mol.
delta Tf = Kf x m => 0.186 = 1.86 x (w2/(M2 x 0.1))
M2 = 1.86 x 0.6 / (0.186 x 0.1) = 1.116/0.0186 = 60 g/mol
The van't Hoff factor of 0.1 M BaCl2 solution is 2.47. Calculate the degree of dissociation.
BaCl2 → Ba2+ + 2Cl-, so n = 3
i = 1 + (n-1) x alpha => 2.47 = 1 + 2 x alpha => alpha = 1.47/2 = 0.735 or 73.5%
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Common mistakes
> - Confusing molarity (per litre of solution) with molality (per kg of solvent). These are different and only molality is temperature-independent.
> - Forgetting to use van't Hoff factor for electrolyte solutions in colligative property problems.
> - Assuming all solutions are ideal — real solutions may show positive or negative deviations from Raoults law.
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Summary
Solutions are homogeneous mixtures. Concentration is expressed as molarity, molality, mole fraction, or percentage. Henrys law governs gas solubility. Raoults law describes vapour pressure of ideal solutions. Colligative properties (vapour pressure lowering, boiling point elevation, freezing point depression, osmotic pressure) depend only on solute particle count, not identity, and are used to find molar masses of unknown substances.