Freezing Point Depression Lab
Freezing Point Depression
Purpose: The purpose of this lab is to demonstrate freezing point depression by dissolving salt into two liquids and monitoring temperature.
Materials: test tube, Thermometer, 10 mL graduated cylinder, 2 micropipettes, Styrofoam cups, 4 pieces plastic wrap, 70% ethyl rubbing alcohol, 91% isopropyl rubbing alcohol, ice, distilled water, rock salt
Procedure: Using the graduated cylinder, measure 2 mL of alcohol and pour it into the test tube. Add 10 mL of distilled water to the test tube also. Put the thermometer into the tube. Cover the tube with plastic wrap. Put ice and salt in the Styrofoam cup to approximately one inch. Put the test tube on top of the ice in the cup. Add salt and ice in the …show more content…
Isopropyl alcohol trial 1 ΔTf = -11°C - 0°C = -11°C Isopropyl alcohol trial 2 ΔTf = -10° C - 0°C = -10°C Ethyl alcohol trial 1 ΔTf = -8° C - 0°C = -8°C Ethyl alcohol trial 2 ΔTf = -7° C - 0°C = -7°C
2. ΔTf = Kfm m = ΔTf / = Kf
Isopropyl alcohol trial 1 m = ΔTf / = Kf = -11°C / -1.86 °C/m = 5.91 m Isopropyl alcohol trial 2 m = ΔTf / = Kf = -10°C / -1.86 °C/m = 5.38 m Ethyl alcohol trial 1 m = ΔTf / = Kf = -8°C / -1.86 °C/m = 4.30 m Ethyl alcohol trial 2 m = ΔTf / = Kf = -7°C / -1.86 °C/m = 3.76 m
4. kg of solvent = volume of solvent x density of solvent kg of solvent = 10 mL x 1 g/mL x 1kg/1000g = 0.01 kg
molality = moles of solute / kg solvent moles of solute = molality x kg solvent
Isopropyl alcohol trial 1 moles of solute = 5.91 m x 0.01 kg = 0.0591 moles Isopropyl alcohol trial 2 moles of solute = 5.38 m x 0.01 kg = 0.0538 moles Ethyl alcohol trial 1 moles of solute = 4.30 m x 0.01 kg = 0.0430 moles Ethyl alcohol trial 2 moles of solute = 3.76 m x 0.01 kg = 0.0376 …show more content…
Assuming there was no human error in following lab procedure, discuss what would be the cause of the percent error.
The first thing is that I used 91% isopropyl alcohol instead of 70%. I took into account that fact in my calculations, but that could have still contributed to the percent error since the experiment was designed for 70% alcohol. Another thing contributing to error could be that some of the alcohol evaporated during the experiment. I would think this would make the temperature not get as low as it should instead of getting too low, however. Another source of error is that possibly the ice and salt did not get to a low enough temperature, but again I would think this would alter the results giving a higher temperature instead of a lower one. Something that would give too low a temperature is if the thermometer bulb was directly on the test tube, and the temperature decreased because of the ice/salt mixture instead of the alcohol/water mixture. This is what probably happened in my experiment. Another thing that would cause the temperature to be too low is contamination in the alcohol, contributing more solute to the solution than
Purpose: The purpose of this lab is to demonstrate freezing point depression by dissolving salt into two liquids and monitoring temperature.
Materials: test tube, Thermometer, 10 mL graduated cylinder, 2 micropipettes, Styrofoam cups, 4 pieces plastic wrap, 70% ethyl rubbing alcohol, 91% isopropyl rubbing alcohol, ice, distilled water, rock salt
Procedure: Using the graduated cylinder, measure 2 mL of alcohol and pour it into the test tube. Add 10 mL of distilled water to the test tube also. Put the thermometer into the tube. Cover the tube with plastic wrap. Put ice and salt in the Styrofoam cup to approximately one inch. Put the test tube on top of the ice in the cup. Add salt and ice in the …show more content…
Isopropyl alcohol trial 1 ΔTf = -11°C - 0°C = -11°C Isopropyl alcohol trial 2 ΔTf = -10° C - 0°C = -10°C Ethyl alcohol trial 1 ΔTf = -8° C - 0°C = -8°C Ethyl alcohol trial 2 ΔTf = -7° C - 0°C = -7°C
2. ΔTf = Kfm m = ΔTf / = Kf
Isopropyl alcohol trial 1 m = ΔTf / = Kf = -11°C / -1.86 °C/m = 5.91 m Isopropyl alcohol trial 2 m = ΔTf / = Kf = -10°C / -1.86 °C/m = 5.38 m Ethyl alcohol trial 1 m = ΔTf / = Kf = -8°C / -1.86 °C/m = 4.30 m Ethyl alcohol trial 2 m = ΔTf / = Kf = -7°C / -1.86 °C/m = 3.76 m
4. kg of solvent = volume of solvent x density of solvent kg of solvent = 10 mL x 1 g/mL x 1kg/1000g = 0.01 kg
molality = moles of solute / kg solvent moles of solute = molality x kg solvent
Isopropyl alcohol trial 1 moles of solute = 5.91 m x 0.01 kg = 0.0591 moles Isopropyl alcohol trial 2 moles of solute = 5.38 m x 0.01 kg = 0.0538 moles Ethyl alcohol trial 1 moles of solute = 4.30 m x 0.01 kg = 0.0430 moles Ethyl alcohol trial 2 moles of solute = 3.76 m x 0.01 kg = 0.0376 …show more content…
Assuming there was no human error in following lab procedure, discuss what would be the cause of the percent error.
The first thing is that I used 91% isopropyl alcohol instead of 70%. I took into account that fact in my calculations, but that could have still contributed to the percent error since the experiment was designed for 70% alcohol. Another thing contributing to error could be that some of the alcohol evaporated during the experiment. I would think this would make the temperature not get as low as it should instead of getting too low, however. Another source of error is that possibly the ice and salt did not get to a low enough temperature, but again I would think this would alter the results giving a higher temperature instead of a lower one. Something that would give too low a temperature is if the thermometer bulb was directly on the test tube, and the temperature decreased because of the ice/salt mixture instead of the alcohol/water mixture. This is what probably happened in my experiment. Another thing that would cause the temperature to be too low is contamination in the alcohol, contributing more solute to the solution than