Purpose
To determine the freezing point dpression of antifreeze solutions by cooling them in an ice-salt bath.
Data Table
Freezing point of tap water (TH2O):
Freezing point of Solution 1 (T1): Freezing point of Solution 2 (T2): |
0.5 degree Celcius
-3.0 degree Celcius -6.0 degree Celcius |
Conclusion
To determine the freezing point depression of antifreeze solutions, we cooled the antifreeze solutions in an ice-salt bath. Using equations such as delta T = Kf x msolute for non-electrolyte solutions and delta T = Kf x msolute x i factor for electrolyte solutions, we were able to determine the freezing point depression for both of the solutions. It is important to remember freezing point depression and boiling point elevation are dependent on colligative properties. The freezing point depression for solution 1 is -3.5 degrees Celcius and solution 2 is -6.5 degrees Celcius. Plugging in the given Kf constant, the molality of solution 1 and 2 were found; solution 1 with 1.88 mol/kg and solution 2 with 3.49 mol/kg. To find the number of grams of antifreeze per 1000 grams of solvent, a proportion was set up to find the mass of the antifreeze; solution 1 had 100 grams of antifreeze while solution 2 had 200 grams of antifreeze. Taking the grams of antifreeze and dividing it by its molality gave the molar mass; solution 1 had a molar mass of 53.16 g/mol while solution 2 had 57.30 g/mol.
Discussion of Theory
A French chemist noted the vapor pressure of a solvent was lowered by adding a solute (Raoult’s Law) and the freezing point of the solution was lowered as well. The equation for finding the change in freezing point of a non-electrolyte solution is:
Delta T is the freezing point depression. Kf is the molal freezing point depression constant of the solvent, and msolute is the molality of the solute in the solution. (Molality is moles of solute over kilograms of solvent.)
Because of colligative properties, freezing point depression and boiling point elevation are dependent on the number solute particles and not of the identity of the particles. The more particles the soluble divides into, the lower the freezing point.
Electrolytes behave in the same way. The equation for finding the change in the freezing point of an electrolyte solution is:
Because of colligative properties, freezing point depression and boiling point elevation are dependent on the number solute particles and not of the identity of the particles. The more particles the soluble divides into, the lower the freezing point.
Electrolytes behave in the same way. The equation for finding the change in the freezing point of an electrolyte solution is:
The Van’t Hoff factor is the I symbol. The I symbol represents the moles of particles in solution over the moles of solute dissolved. It’s basically how many ions a compound splits into. For example, NaCl would split into one Na and one Cl ion; thus, the I factor would equal 2. Another example, BaCl2 would split into one Ba ion and two Cl ions; thus, the I factor would be 3. For strong acids/bases, almost all of them has an I factor of 2 except for H2SO4 (which is 2+ because HSO4- is a weak acid) If it is a weak acid/base, the solute stays together but a small part of it dissociates; therefore, the i solute value is a little more than 1, or 1+.
Sources of Error
The water used in this lab was tap water instead of distilled water. Tap water has loads of impurities which can conflict with the Van’t Hoff factor and increase the freezing point depression. The freezing point should have been lower than 0 degrees Celcius with the water used in this lab because it was not pure. In this lab, not all of the NaCl was able to get out of the cup. This could have skewed the data of the freezing point to have a higher temperature. In the beginning of the lab, when finding the freezing point of the tap water, the water just froze instead of forming ice crystals/slush. There probably wasn’t enough stirring or something for that to have happened. The freezing point would have been higher. On the second solution, it did not slush so more water and salt was added (unmeasured) and it still did not slush, so data was taken from another group.
Critical Thinking Analysis
3. The water used in this lab was tap water instead of distilled water. Tap water has loads of impurities which can conflict with the Van’t Hoff factor and increase the freezing point depression. Using pure water such as Fiji water would fix this issue. In this lab, not all of the NaCl was able to get out of the cup. This could have skewed the data of the freezing point to have a higher temperature. Perhaps water could have been in the cup that held the NaCl and poured into the 250mL beaker.
Critical Thinking Applications
1. Freezing point depression can’t be used for substances not soluble in water. Insoluble substances don’t dissolve in water since they don’t break up into smaller particles. This means they have an I factor of one. Since molality is the moles of solute divided by the mass in kilograms of the solvent, and insoluble substances don’t dissove, they have a molality of zero in which the change in freezing point is zero as well. Freezing point depression can not be used for insoluble substances in water.
2. One molal solution of the substance (NH4)3PO4 would have an impact on the freezing point depression of water. The freezing point depression equation is delta Tf =iKf*Msolute. The Van’t Hoff factor in this situation is 4 because (NH4)3PO4 would break up into 4 ions. The Kf constant is unique to the solvent. With the molality being 1, it doesn’t have a huge impact on changing the freezing point, but the Van’t Hoff factor increases the change in the freezing point temperature since I=4. If the molality was >1, it would be a greater change with a lower freezing point.
3. The assumption is that the density of distilled water is 1g/1mL. It was used to convert the volume of water to grams of water which was used to find the grams of the antifreeze (per 1000 grams of solvent) and the molar mass.
4. Soluble substances have molality as opposed to insoluble substances, so the molar mass determination would be practical for other substances soluble in water. After finding the change in freezing point, the molality can be figured out. The solvent and the volume of water would be converted to grams. A proportion of the grams of solute would then be divided by the grams of solvent used in the experiment equal to the amount of grams of the solute per 1000 grams of water; thus, the amount of grams of solute per 1000 g can be solved. Taking this value (units of grams) and dividing it by the molality (units of mol/kg), the kg units cancels out, leaving only grams over mole (molar mass units). This method does not only apply to this antifreeze investigation.
2. One molal solution of the substance (NH4)3PO4 would have an impact on the freezing point depression of water. The freezing point depression equation is delta Tf =iKf*Msolute. The Van’t Hoff factor in this situation is 4 because (NH4)3PO4 would break up into 4 ions. The Kf constant is unique to the solvent. With the molality being 1, it doesn’t have a huge impact on changing the freezing point, but the Van’t Hoff factor increases the change in the freezing point temperature since I=4. If the molality was >1, it would be a greater change with a lower freezing point.
3. The assumption is that the density of distilled water is 1g/1mL. It was used to convert the volume of water to grams of water which was used to find the grams of the antifreeze (per 1000 grams of solvent) and the molar mass.
4. Soluble substances have molality as opposed to insoluble substances, so the molar mass determination would be practical for other substances soluble in water. After finding the change in freezing point, the molality can be figured out. The solvent and the volume of water would be converted to grams. A proportion of the grams of solute would then be divided by the grams of solvent used in the experiment equal to the amount of grams of the solute per 1000 grams of water; thus, the amount of grams of solute per 1000 g can be solved. Taking this value (units of grams) and dividing it by the molality (units of mol/kg), the kg units cancels out, leaving only grams over mole (molar mass units). This method does not only apply to this antifreeze investigation.