Lyophilic sols are an important class of colloidal systems commonly studied in physical chemistry. The topic Lyophilic sols properties examples preparations is frequently asked in school and competitive examinations because it connects theory with real-life applications. Lyophilic sols are formed when the dispersed phase has a strong affinity for the dispersion medium, leading to highly stable colloidal systems.

To clearly understand Lyophilic Sols Properties Examples Preparations, it is essential to begin with their basic definition. Lyophilic sols are colloidal solutions in which the dispersed particles are strongly solvated by the solvent molecules. The word “lyophilic” literally means “liquid loving,” indicating the strong attraction between the sol particles and the dispersion medium.

Properties of Lyophilic Sols

One of the most important aspects of Lyophilic sols properties examples preparations is their exceptional stability. Lyophilic sols are highly stable because solvation forms a protective layer around the colloidal particles. This prevents aggregation and coagulation.

Another key point in Lyophilic sols properties examples preparations is reversibility. Lyophilic sols are reversible in nature. If the solvent is removed by evaporation, the sol can be re-formed simply by adding the solvent again.

Viscosity is also a notable feature discussed under Lyophilic sols properties examples preparations. These sols increase the viscosity of the dispersion medium significantly due to strong interaction between dispersed phase and solvent.

Lyophilic sols do not exhibit the Tyndall effect prominently, which is another important property covered in Lyophilic sols properties examples preparations. This is because the particles are heavily solvated and closely resemble true solutions.

Additionally, lyophilic sols are less sensitive to electrolytes. Even high concentrations of electrolytes are often required to cause coagulation, a concept emphasized in Lyophilic sols properties examples preparations.

Examples of Lyophilic Sols

When studying Lyophilic sols properties examples preparations, common examples help in better conceptual clarity. Typical examples include sols of starch, gelatin, gum, proteins, and cellulose derivatives.

Starch sol in water is a classic example highlighted in Lyophilic sols properties examples preparations. Gelatin in water is another widely cited example, especially in biological and pharmaceutical contexts.

Proteins dissolved in water form lyophilic sols, which is why the topic Lyophilic sols properties examples preparations is also relevant in biochemistry and medicine. Rubber dissolved in organic solvents is another example that demonstrates lyophilic behavior.

Preparation of Lyophilic Sols

The preparation methods form a crucial part of Lyophilic sols properties examples preparations. Lyophilic sols are generally easy to prepare compared to lyophobic sols.

The most common method discussed in Lyophilic sols properties examples preparations is direct dissolution. In this method, the dispersed phase is simply dissolved in a suitable dispersion medium under proper conditions.

Heating and stirring often assist the preparation process, which is another detail emphasized in Lyophilic sols properties examples preparations. For example, gelatin sol is prepared by heating gelatin in water with continuous stirring.

Unlike lyophobic sols, no special stabilizing agents are required during preparation. This simplicity is a defining feature explained under Lyophilic sols properties examples preparations.

Importance and Applications

Understanding Lyophilic sols properties examples preparations is essential due to their wide applications. They are used in food industries (gelatin desserts), pharmaceuticals (drug delivery systems), cosmetics, and biological systems.

The stability and reversibility discussed in Lyophilic sols properties examples preparations make these sols highly useful in industrial formulations and research applications

MCQs on Lyophilic Sols Properties Examples Preparations with Answers-

1. Observe the following abbreviations:
   Πobs\Pi_{obs}Πobs​ = observed colligative property, Πcal\Pi_{cal}Πcal​ = theoretical colligative property (normal behavior). Van’t Hoff factor (i) is given by

A) i=Πobs×Πcali=\Pi_{obs}\times \Pi_{cal}i=Πobs​×Πcal​
B) i=Πobs+Πcali=\Pi_{obs}+\Pi_{cal}i=Πobs​+Πcal​
C) i=Πobs−Πcali=\Pi_{obs}-\Pi_{cal}i=Πobs​−Πcal​
D) i=ΠobsΠcali=\dfrac{\Pi_{obs}}{\Pi_{cal}}i=Πcal​Πobs​​

Answer: D

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2. Lyophilic sols are more stable than lyophobic sols because the particles:

A) are positively charged
B) are negatively charged
C) are solvated
D) repel each other

Answer: B

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3. The molal elevation constant for water is 0.52. What will be the boiling point of 2 molal sugar solution at 1 atm? (B.P. of pure water = 100°C)

A) 101.04°C
B) 100.26°C
C) 100.52°C
D) 99.74°C

Answer: A

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4. Molal solution means 1 mole of solute present in:

A) 1000 g of solvent
B) 1 L of solvent
C) 1 L of solution
D) 1000 g of solution

Answer: C

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5. Which of the following shows maximum depression in freezing point?

A) K₂SO₄
B) NaCl
C) Urea
D) Glucose

Answer: A

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6. A solution of 4.5 g of a pure non-electrolyte in 100 g of water was found to freeze at 0.465°C. The molecular weight of the solute is closest to ( Kf=1.86K_f = 1.86Kf​=1.86 )

A) 13.5
B) 172
C) 90
D) 180

Answer: D

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7. If the elevation in boiling point of a solution of 10 g of solute (mol. wt. = 100) in 100 g of water is ΔTb\Delta T_bΔTb​, the ebullioscopic constant of water is:

A) 10
B) 100Tb100T_b100Tb​
C) ΔTb\Delta T_bΔTb​
D) ΔTb10\dfrac{\Delta T_b}{10}10ΔTb​​

Answer: C

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8. Equimolar solution in the same solvent have:

A) different boiling point and different freezing point
B) same boiling point and same freezing point
C) same freezing point but different boiling point
D) same boiling point but different freezing point

Answer: B

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9. A pressure cooker reduces cooking time for food because:

A) heat is more evenly distributed
B) boiling point of water is increased
C) higher pressure crushes the food
D) chemical changes are helped by rise in temperature

Answer: B

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10. Calculate the quantity of CO₂ required to prepare 1 L of soda water when packed under 2 atm of CO₂. (Henry’s law constant for CO₂ = 1.67×1081.67\times10^81.67×108 Pa)

A) 5.9 g
B) 12.1 g
C) 6.7 g
D) 2.9 g

Answer: D

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11. Henry’s law constant for solubility of N₂ in water at 298 K is 1×1051\times10^51×105 atm. Mole fraction of N₂ in air is 0.8. Moles of N₂ dissolved in 10 moles of water at 298 K and 5 atm is

A) 4×10−44\times10^{-4}4×10−4
B) 8×10−48\times10^{-4}8×10−4
C) 5×10−45\times10^{-4}5×10−4
D) 4×10−34\times10^{-3}4×10−3

Answer: A

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12. Henry’s law is valid for
    (i) NH₃ in water (ii) O₂ in unsaturated blood (iii) O₂ in water (iv) CO₂ in water

A) (i, ii)
B) (ii, iii)
C) (iv, iii)
D) (ii, iv)

Answer: C

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13. The vapour pressure of two liquids X and Y are 80 and 60 Torr. Vapour pressure of ideal solution obtained by mixing 3 moles X and 2 moles Y is

A) 68 Torr
B) 140 Torr
C) 48 Torr
D) 72 Torr

Answer: D

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14. Which option is inconsistent for Raoult’s law?

A) Volume(solvent)+Volume(solute)=Volume(solution)
B) Heat of dilution = 0
C) Solute does not undergo association
D) Solute undergoes dissociation

Answer: D

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15. 18 g of glucose is added to 178.2 g of water. Vapour pressure of water (torr) for this solution is

A) 76
B) 752.4
C) 759.4
D) 76.4

Answer: B

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16. For dilute solutions, Raoult’s law states that

A) lowering of vapour pressure = mole fraction of solute
B) relative lowering = mole fraction of solvent
C) relative lowering of vapour pressure of solvent = mole fraction of solute
D) vapour pressure of solution = vapour pressure of solvent

Answer: C

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17. Relative lowering of vapour pressure is 0.0125. Molality of solution is

A) 0.7
B) 0.5
C) 0.6
D) 0.4

Answer: A

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18. Vapour pressure of benzene = 640 mmHg. 2.175 g solute added to 39.08 g benzene. Vapour pressure becomes 600 mmHg. Molecular weight of solute is

A) 49.5
B) 59.6
C) 69.6
D) 89.62

Answer: C

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19. Vapour pressure decreases by 10 mmHg when solute added. Mole fraction of solute = 0.2. What mole fraction of solvent is needed for decrease to be 20 mmHg?

A) 0.4
B) 0.6
C) 0.2
D) 0.1

Answer: B

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20. Mixture showing positive deviation from Raoult’s law is

A) Ethanol + acetone
B) Benzene + toluene
C) Acetone + chloroform
D) Chloroform + bromobenzene

Answer: A

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21. Value of Henry’s constant KhK_hKh​

A) no effect by temperature
B) decreases with increase in temperature
C) increases with increase in temperature
D) first decreases then increases

Answer: C

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22. Pick the correct statement

A) Relative lowering of vapour pressure is independent of T
B) Osmotic pressure always depends on nature of solute
C) Elevation in boiling point is independent of solvent nature
D) Lowering of freezing point ∝ molar concentration of solute

Answer: A

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23. 12.2 g benzoic acid in 100 g water gives freezing point -0.93°C. Kf=1.86K_f=1.86Kf​=1.86. Number (n) of benzoic acid molecules associated (100% association) is

A) 1
B) 2
C) 3
D) 4

Answer: B

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24. Freezing point of solution containing 25 g ethanol in 1000 g water is ( Kf=1.86K_f=1.86Kf​=1.86 )

A) -0.25°C
B) -0.50°C
C) -1.5°C
D) -1.0°C

Answer: D

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25. 224 mL SO₂(g) at 298 K, 1 atm passed through 100 mL 0.1 M NaOH. Solute formed dissolved in 36 g water. Lowering of vapour pressure ( PH2O=24P_{H2O}=24PH2O​=24 mmHg ) is x×10−2x\times10^{-2}x×10−2 mmHg. x =

A) 18
B) 24
C) 36
D) 48

Answer: A

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26. Pure vapour pressures: A = 450 mmHg, B = 700 mmHg at 350 K. Total vapour pressure of mixture = 600 mmHg. Composition in solution is

A) xA=0.4,  xB=0.6x_A=0.4,\;x_B=0.6xA​=0.4,xB​=0.6
B) xA=0.6,  xB=0.4x_A=0.6,\;x_B=0.4xA​=0.6,xB​=0.4
C) xA=0.3,  xB=0.7x_A=0.3,\;x_B=0.7xA​=0.3,xB​=0.7
D) xA=0.7,  xB=0.3x_A=0.7,\;x_B=0.3xA​=0.7,xB​=0.3

Answer: A

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27. Henry’s law constant for N₂ in water at 298 K = 1.0×1051.0\times10^51.0×105 atm. Mole fraction of N₂ in air = 0.8. Moles of N₂ dissolved in 10 moles water at 5 atm is

A) 4.0×10−44.0\times10^{-4}4.0×10−4
B) 4.0×10−54.0\times10^{-5}4.0×10−5
C) 5.0×10−45.0\times10^{-4}5.0×10−4
D) 5.0×10−55.0\times10^{-5}5.0×10−5

Answer: A

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28. Vapour pressure of solution at 45°C with benzene:octane = 3:2 (benzene = 280 mmHg, octane = 420 mmHg, ideal) is

A) 350 mmHg
B) 160 mmHg
C) 168 mmHg
D) 336 mmHg

Answer: D

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29. At 363 K, vapour pressure of A = 21 kPa, B = 18 kPa. Mix 1 mole A + 2 moles B (ideal). Vapour pressure (nearest integer) is

A) 20 kPa
B) 19 kPa
C) 22 kPa
D) 18 kPa

Answer: B

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30. 180 g water + 10 g each of solutes A, B, C. Relative lowering of vapour pressure order given molar masses: A=100, B=200, C=10000 is

A) A > B > C
B) B > C > A
C) A > C > B
D) C > A > B

Answer: A

Conclusion

In summary, Lyophilic sols properties examples preparations form a core topic in colloid chemistry. Their high stability, reversibility, strong solvation, and ease of preparation distinguish them from lyophobic sols. With numerous real-life examples and straightforward preparation methods, mastering Lyophilic sols properties examples preparations helps aspirants build a strong foundation in physical chemistry and perform confidently in exams.