To predict and measure the freezing point depression caused by a solute in a solvent.
A solution is a homogeneous mixture of two or more substances. For the common two-component solution, the substance present in the major proportion is called the solvent and that in the minor proportion is called the solute. Solvent properties are often changed by the presence of the solute, generally in proportion to the amount of solute concentration. Dissolving copper sulfate in water causes the resulting solution to have a blue color. Dissolving sodium hydroxide or ammonia in water produces a basic solution. Adding molasses or honey to water produces a solution more viscous than water.There are, however, a few properties of a solution that are affected by the concentration of particles the solute contributes to the solution regardless of their identity. These are called colligative properties and include changes in osmotic pressure, vapor pressure lowering, boiling point elevation and freezing point depression.The change in the freezing or boiling point of a solvent when a solute is added is proportional to the colligative molality (mc) of the solution.
Sample Freezing Point Depression problem.....
31.65 g of sodium chloride is added to 220.0 mL of water at 34 °C. How will this affect the freezing point of the water?
Assume the sodium chloride completely dissociates in the water.
Given: density of water at 35 °C = 0.994 g/mL
Kf water = 1.86 °C kg/mol
To find the temperature change elevation of a solvent by a solute, use the equation:
ΔT = iKfm
ΔT = Change in temperature in °C
i = van 't Hoff factor
Kf = molal freezing point depression constant or cryoscopic constant in °C kg/mol
m = molality of the solute in mol solute/kg solvent.
Step 1 Calculate the molality of the NaCl
molality (m) of NaCl = moles of NaCl/kg water
From the periodic table
atomic mass Na = 22.99
atomic mass Cl = 35.45
moles of NaCl = 31.65 g x 1 mol/(22.99 + 35.45)
moles of NaCl = 31.65 g x 1 mol/58.44 g
moles of NaCl = 0.542 mol
kg water = density x volume
kg water = 0.994 g/mL x 220 mL x 1 kg/1000 g
kg water = 0.219 kg
mNaCl = moles of NaCl/kg water
mNaCl = 0.542 mol/0.219 kg
mNaCl = 2.477 mol/kg
Step 2 Determine the van 't Hoff factor
The van 't Hoff factor, i, is a constant associated with the amount of dissociation of the solute in the solvent. For substances which do not dissociate in water, such as sugar, i = 1. For solutes that completely dissociate into two ions, i = 2. For this example NaCl completely dissociates into the two ions, Na+ and Cl-. Therefore, i = 2 for this example.
Step 3 Find ΔT
ΔT = iKfm
ΔT = 2 x 1.86 °C kg/mol x 2.477 mol/kg
ΔT = 9.21 °C
Adding 31.65 g of NaCl to 220.0 mL of water will lower the freezing point by 9.21 °C.
Brief presentation about freezing point depression as a colligative property
Source: adapted from GT Chemistry Curriculum 2010, Baltimore County Public Schools