2. Freezing Point Depression

4. When the rate of freezing is the same as the rate of melting, the amount of ice and the amount of water won't change. The ice and water are said to be in dynamic equilibrium with each other.

5. Here's the same container with the water at 0°C, only this time the water contains salt molecules. Adding salt disrupts the equilibrium. The salt molecules dissolve, but do not attach easily to the solid ice. There are fewer water molecules; some of the water has been replaced by salt. The number of water molecules able to be captured by the ice decreases, so the rate of freezing slows.

6. The rate of melting of the ice is unchanged by the salt so melting is now occurring faster than freezing. So the ice eventually melts. The temperature drops so that the rate of freezing can increase to reach equilibrium. In this example, the new freezing/melting point is reached at -4°C. The higher the concentration of salt, the lower the freezing point drops.

10. Calculate the change in freezing point ( Δ T f ) relative to the pure solvent using the equation: Δ T f = i K f m where K f is the freezing point depression constant for the solvent (1.86°C·kg/mol for water) m is the number of moles of solute in solution per kilogram of solvent and i is the number of ions present per formula unit (e.g., i = 2 for NaCl).

11. 1.60 g of naphthalene (C 10 H 8 ) is dissolved in 20.0 g of benzene. The freezing point of pure benzene is 5.5 o C, and the freezing point of the mixture is 2.8 o C. What is the molal freezing point depression constant, K f of benzene?

12. Step 1: Calculate the freezing point depression of benzene. T f = (Freezing point of pure solvent) - (Freezing point of solution) (5.5 o C) - (2.8 o C) = 2.7 o C Step 2 : Calculate the molal concentration of the solution. molality = moles of solute / kg of solvent moles of naphthalene = (1.60 g) (1 mol / 128 g) = 0.0125 mol naphthalene molality of solution = (0.0125 mol) / (0.0200 kg) = 0.625 m

13. Step 3: Calculate K f of the solution. T f = (K f ) (m) (2.7 o C) = (K f ) (0.625 m) K f = 4.3 o C/m

14. References Read more: Colligative Properties - Chemistry Encyclopedia - water, uses, examples, gas, number, equation, salt, property, Freezing Point Depression, Boiling Point Elevation, Vapor Pressure Lowering, Osmotic Pressure http://www.chemistryexplained.com/Ce-Co/Colligative- Properties.html#ixzz1ESZg9En9Colligative Properties - Chemistry Encyclopedia - water, uses, examples, gas, number, equation, salt, property, Freezing Point Depression, Boiling Point Elevation, Vapor Pressure Lowering, Osmotic Pressure http://www.chemistryexplained.com/Ce-Co/Colligative- Properties.html#ixzz1ESZg9En9 http://www.worsleyschool.net/science/files/saltandfreezi ng/ofwater.html http://www.worsleyschool.net/science/files/saltandfreezi ng/ofwater.html http://chemed.chem.purdue.edu/genchem/probsolv/colliga tive/kf1.3.html http://chemed.chem.purdue.edu/genchem/probsolv/colliga tive/kf1.3.html Image: http://faculty.clintoncc.suny.edu/faculty/michael.gregory/ files/bio%20101/bio%20101%20lectures/chemistry/chemist r.htm http://faculty.clintoncc.suny.edu/faculty/michael.gregory/ files/bio%20101/bio%20101%20lectures/chemistry/chemist r.htm