Part 1: Multiple choice
A solution is
In natural aquatic environments the solvent is usually
Colligative properties of a solution depend on
Which is NOT a colligative property?
The number of colligative properties named in the book is
Dynamic equilibrium between a liquid and its vapour in a closed vessel means
The pressure exerted by the vapour above a liquid in equilibrium is its
Dissolving a non-volatile solute in water
The decrease in vapour pressure is directly proportional to
A liquid boils when its vapour pressure becomes equal to
The boiling point of pure water under standard pressure is
At a high mountain top, water boils at a temperature
A pressure cooker cooks food faster because inside it
The boiling point of a pure liquid under constant pressure can be used as
Compared with pure water, an aqueous salt solution boils
Compared with pure water, an aqueous salt solution freezes
Salt is spread on roads in cold countries to
Osmosis is the movement of
A semi-permeable membrane allows
Water moves by osmosis toward the solution that has
A red blood cell placed in distilled water will
A plant cell placed in a very concentrated sugar solution shows
A solution with the same osmotic pressure as the cell sap is
Which 0.1 mol/kg solution has the lowest freezing point?
Which 0.1 mol/kg solution has the highest vapour pressure?
Drinking sea water makes a person more thirsty because
Reverse osmosis desalination works by applying
Salting meat or fish preserves it because bacteria
Which statement about NaCl and glucose (equal moles in equal water) is correct?
A non-volatile solute is one that
If an external pressure on the surface of a liquid is increased, its boiling point
Sea water freezes at about
In the U-tube experiment with pure water on one side and sugar solution on the other, after some time the level
Which of the following solutions is isotonic with body fluids?
Plant roots absorb water from soil by osmosis because the cell sap in root hairs is
Over-fertilising a field may cause plants to wilt because
When the dissolved salt concentration of an enclosed lake increases due to evaporation, the vapour pressure of the lake water
The rate of evaporation and condensation are equal in
The ice of salty sea water that forms in polar seas leaves the salt behind, so the water below becomes
A solution containing 0.2 mol of particles per kg of water compared with one containing 0.4 mol of particles per kg has
Part 2: Written questions and problems
1. [Give reasons] Give a reason: the vapour pressure of a solution of a non-volatile solute is lower than that of the pure solvent.
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Solute particles occupy part of the surface and attract water molecules, so fewer water molecules can evaporate.
2. [Give reasons] Give a reason: the boiling point of salt solution is higher than that of pure water.
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Its vapour pressure is lower, so it must be heated to a higher temperature before the vapour pressure equals the external pressure.
3. [Give reasons] Give a reason: salt is scattered on frozen roads.
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The salt lowers the freezing point of water, preventing ice formation and skidding.
4. [Give reasons] Give a reason: water boils at less than 100 °C at the top of a high mountain.
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Atmospheric pressure is lower, so the vapour pressure needed for boiling is reached at a lower temperature.
5. [Give reasons] Give a reason: food cooks faster in a pressure cooker.
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The raised pressure increases the boiling point (about 120 °C), so the food is cooked at a higher temperature.
6. [Give reasons] Give a reason: the boiling point of a pure liquid is a good indicator of its purity.
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It is constant for a pure liquid at a fixed pressure while impurities change it.
7. [Give reasons] Give a reason: a red blood cell swells and bursts in distilled water.
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The cell contents are more concentrated; water enters by osmosis and the thin membrane cannot withstand the volume.
8. [Give reasons] Give a reason: a plant cell does not burst in distilled water.
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The strong cellulose cell wall resists the expansion; the cell becomes turgid instead.
9. [Give reasons] Give a reason: pickling and salting preserve foods.
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The high salt concentration draws water out of microbes by osmosis, so they cannot grow.
10. [Give reasons] Give a reason: equal masses of NaCl and sugar do not have equal colligative effects.
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Colligative effects depend on the number of particles; NaCl gives two ions per formula unit and has a lower molar mass than sugar, so it provides many more particles.
11. [Give reasons] Give a reason: drinking sea water cannot satisfy thirst.
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Its salt concentration is higher than that of body fluids; water moves out of cells and the kidneys need extra water to remove the salt.
12. [Give reasons] Give a reason: the freezing point of sea water is below 0 °C.
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Dissolved salts hinder formation of the ice lattice.
13. [Give reasons] Give a reason: plants wilt when irrigated with very salty water.
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The soil solution becomes more concentrated than the root sap; water leaves the root cells by osmosis.
14. [Give reasons] Explain how dissolved salts in the ocean help to create vertical ocean currents.
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Salinity (and temperature) changes density; denser salty or cold water sinks and lighter water rises, creating vertical currents.
15. [Give reasons] Explain what is meant by saying that the colligative properties do not depend on the type of solute.
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Equal numbers of dissolved particles of different non-volatile solutes give the same effect in a given amount of solvent.
Part C — Define / compare
16. [Define] Define colligative properties.
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Properties of a solution that depend on the number of molecules or ions of a non-volatile solute and not on its type.
17. [Define] Define the vapour pressure of a liquid.
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The pressure exerted by the vapour in dynamic equilibrium with its liquid at a given temperature in a closed vessel.
18. [Define] Define the boiling point of a liquid.
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The temperature at which the vapour pressure of the liquid equals the external pressure on its surface.
19. [Define] Define osmosis.
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Movement of water from a dilute solution to a more concentrated one through a semi-permeable membrane.
20. [Define] Define osmotic pressure.
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The pressure arising in a solution due to a difference in solute concentration, which makes water move toward the solution of higher concentration.
21. [Compare] Compare hypotonic, isotonic and hypertonic solutions with respect to a cell.
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Hypotonic: more dilute than the cell, water enters. Isotonic: equal, no net movement. Hypertonic: more concentrated, water leaves.
22. [Compare] Compare the effect of placing an animal cell and a plant cell in distilled water.
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Animal cell: swells and bursts (no cell wall). Plant cell: becomes turgid, protected by the cell wall.
23. [Compare] Compare the effect of a non-volatile solute on the boiling point and the freezing point of water.
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It raises the boiling point and lowers the freezing point; both effects are proportional to the number of dissolved particles.
24. [Compare] Compare 0.1 mol NaCl and 0.1 mol glucose in 1 kg water in their effect on freezing point.
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NaCl (0.2 mol of particles) lowers it by 0.372 °C; glucose (0.1 mol) lowers it by 0.186 °C: NaCl has twice the effect.
25. [Compare] Compare pure liquid and solution regarding vapour pressure and boiling point.
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Pure liquid: higher vapour pressure, lower constant boiling point. Solution: lower vapour pressure, higher boiling point.
Part D — Problems
26. [Problem] 171 g of sucrose (M = 342) is dissolved in 162 g of water at 25 °C. Using P° = 23.8 mmHg, find the vapour pressure of the solution.
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≈ 22.5 mmHg.
n(sucrose) = 0.5; n(water) = 9; χ(water) = 9/9.5 = 0.947; P = 0.947 × 23.8 = 22.5.
27. [Problem] Find the freezing point of a solution containing 0.5 mol glucose in 1 kg of water.
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−0.93 °C.
ΔTf = 1.86 × 0.5 = 0.93.
28. [Problem] Find the boiling point of a solution of 1 mol CaCl₂ in 1 kg water, assuming complete dissociation.
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101.54 °C.
i = 3; ΔTb = 3 × 0.512 × 1 = 1.536 °C.
29. [Problem] 29.25 g NaCl is dissolved in 500 g of water. Find the freezing point (ideal).
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−3.72 °C.
n = 0.5 mol; m = 1.0 mol/kg; ΔTf = 2 × 1.86 × 1.0 = 3.72.
30. [Problem] Ethylene glycol (M = 62) is added to water as antifreeze: 310 g in 1 kg of water. Find the freezing point.
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−9.3 °C.
n = 5 mol; m = 5; ΔTf = 1.86 × 5 = 9.3.
31. [Problem] Calculate the osmotic pressure of 0.1 mol/L glucose at 25 °C.
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≈ 2.45 atm.
π = MRT = 0.1 × 0.0821 × 298 = 2.45.
32. [Problem] Calculate the osmotic pressure of 0.1 mol/L CaCl₂ at 25 °C (complete dissociation).
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≈ 7.34 atm.
i = 3; π = 3 × 2.45 = 7.34.
33. [Problem] A glucose solution has osmotic pressure 4.9 atm at 300 K. Find its molarity.
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≈ 0.20 mol/L.
M = π/(RT) = 4.9/(0.0821 × 300) = 0.199.
34. [Problem] What mass of urea (M = 60) should be dissolved in 1 kg of water to lower the freezing point by 1.86 °C?
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60 g.
ΔTf = 1.86 requires m = 1 mol/kg of a non-ionising solute.
35. [Problem] 0.25 mol sucrose is dissolved in 500 g water. Find the boiling point.
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100.26 °C.
m = 0.5; ΔTb = 0.512 × 0.5 = 0.256.
36. [Problem] Rank 0.1 mol/kg glucose, NaCl and CaCl₂ by freezing point and give each value.
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CaCl₂ −0.558 °C < NaCl −0.372 °C < glucose −0.186 °C.
ΔTf = i × 1.86 × 0.1 with i = 3, 2, 1.
37. [Problem] Sea water contains about 35 g NaCl per kg. Estimate (ideal-solution) its freezing point.
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≈ −2.2 °C (the real value, −1.9 °C, is a little higher because the solution is not ideal).
n = 35/58.5 = 0.598 mol; ΔTf = 2 × 1.86 × 0.598 = 2.2.
38. [Problem] A cooker raises the pressure so water boils at 120 °C. By how many kelvin is the boiling point above normal, and what is it in kelvin?
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20 K higher; 393 K.
120 − 100 = 20; 120 + 273 = 393.
39. [Problem] 14.2 g of sodium sulfate (Na₂SO₄ = 142) is dissolved in 1 kg of water. Find the freezing point (ideal, complete dissociation).
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−0.56 °C.
n = 0.1 mol; i = 3 (2Na⁺ + SO₄²⁻); ΔTf = 3 × 1.86 × 0.1 = 0.558.
40. [Problem] By what percentage is the vapour pressure of water lowered when mole fraction of water is 0.98?
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2 %.
P = 0.98 P°, a lowering of 0.02 P°.