Depression of Freezing Point - UNSOLVED PRACTICE SET

Q7. Define depression of freezing point. Derive the expression ΔTf = Kf · m for a dilute solution.

Q8. Calculate the freezing point of a solution containing 6.0 g of urea in 200 g of water. (Kf for water = 1.86 K·kg/mol, freezing point of pure water = 0°C)

[Given: Molar mass of urea = 60 g/mol]

Q9. Why is camphor preferred as a solvent in the Rast method for determining molar mass? What makes it special

Q10. Explain why the freezing point of a solution is lower than that of the pure solvent at the molecular level.

Q11. Your father adds antifreeze (ethylene glycol) to the car radiator before winter in Shimla. Explain how this prevents the coolant from freezing.

Q12. Why does seawater remain liquid at temperatures below 0°C? Calculate the freezing-point depression for seawater containing 35 g/L NaCl.

Q13. Discuss depression of freezing point in detail:

(a) Definition and explanation using vapour pressure curves

(b) Derivation of the relationship ΔTf = Kf · m

(c) Definition and significance of cryoscopic constant (Kf)

(d) Determination of molar mass using freezing point depression

(e) Experimental methods: Beckmann's method and Rast's method

(f) Limitations and sources of error

Q14. Explain the factors affecting depression of freezing point:

(a) Nature of solvent — how Kf varies with solvent properties

(b) Concentration of solute — linear relationship with molality

(c) Nature of solute — electrolytes vs. non-electrolytes (van't Hoff factor)

(d) Association and dissociation of solute particles

(e) Comparison of experimental and calculated values for different solutes

(f) Why freezing point depression is more sensitive than boiling point elevation for molar mass determination

Q15. Freezing point depression has numerous practical applications in India. Discuss:

(a) Why salt is spread on icy roads in Kashmir and Himachal Pradesh during winter

(b) The use of ethylene glycol antifreeze in vehicles across India's diverse climate zones

(c) How freezing point depression is used to detect water adulteration in milk (cryoscopy in dairy industry)

(d) The role of freezing point depression in making smooth ice cream (preventing large ice crystal formation)

Q16. The freezing point depression constant for benzene is 5.12 K·kg/mol. A solution of 2.56 g of naphthalene in 100 g of benzene freezes at 4.98°C. (Pure benzene freezes at 5.50°C).

(a) Calculate the depression of freezing point.

(b) Calculate the molality of the solution.

(c) Calculate the molar mass of naphthalene and compare with the theoretical value.

Q17. A car radiator in Leh, Ladakh must be protected against freezing to -30°C.

(a) Calculate the minimum mass of ethylene glycol (C₂H₆O₂) that must be added to 5.0 kg of water to prevent freezing at -30°C. (Kf for water = 1.86 K·kg/mol)

(b) Calculate the boiling point of this mixture. (Kb for water = 0.512 K·kg/mol)

(c) If the radiator capacity is 6.0 L and the density of the mixture is 1.05 g/mL, verify whether your calculated amount fits in the radiator.

[Given: Molar mass of ethylene glycol = 62 g/mol]

Q18. The freezing point of milk is used to detect water adulteration. Pure cow milk freezes at approximately -0.540°C.

(a) A milk sample freezes at -0.486°C. Calculate the percentage of water added as adulterant.

(b) If the milk cooperative tests 1000 samples daily and rejects those with >5% water, calculate how many samples would be rejected if the average adulteration is 8%.

(c) Explain why freezing point depression is preferred over density measurement for detecting water adulteration.

[Given: Kf for water = 1.86 K·kg/mol, assume milk is primarily water with dissolved solids equivalent to 0.3 molal solution]