Chem 112 Lab Report 14 Questions: Molar Mass of a Solid
Angela Pierre
Chem 112 Lab
Due: 03/05/2015
Lab Report 14 Questions
1. Part A.3. Some of the cyclohexane solvent vaporized during the temperature versus time measurement. Will this loss of cyclohexane result in its freezing point being recorded as too high, too low, or unaffected? Explain.
-The freezing point will be higher because the loss of solvent means that the molarity of solute is too high. The melting point will be lower than expected.
5. Part B.2. The solute dissociates slightly in the solvent. How will the slight dissociation affect the reported molar mass of the solute- too high, too low, or unaffected? Explain.
-The solute dissociating slightly will cause you to calculate a molar mass that is too small if you neglect the dissociation by assuming i=1 instead of the correct value of i, which is >1.
6. Part B.3. The temperature versus time data plot shows no change in your temperature at the freezing point for a pure solvent; however, the temperature at the freezing point for a solution steadily decreases until the solution has completely solidified. Account for this decreasing temperature.
- The freezing point of the solution is always lower than the freezing point of the pure solvent because when a non-volatile solute is dissolved in a solvent, the vapor pressure of the solvent is lowered. A depression of the solvent freezing point occurs.
7. Part C.1. Interpretation of the data plots consistently shows that the freezing points of three solutions are too high. As a result of this “misreading of the data,” will the reported molar mass of the solute be too high, too low, or unaffected? Explain.
-The change in temperature is suppose to increase because the freezing point from the cooling curves increases causing the molar mass to be lower. If three solutions are too high then the molar mass will be recorded as high.
Chem 112 Lab
Due: 03/05/2015
Lab Report 14 Questions
1. Part A.3. Some of the cyclohexane solvent vaporized during the temperature versus time measurement. Will this loss of cyclohexane result in its freezing point being recorded as too high, too low, or unaffected? Explain.
-The freezing point will be higher because the loss of solvent means that the molarity of solute is too high. The melting point will be lower than expected.
5. Part B.2. The solute dissociates slightly in the solvent. How will the slight dissociation affect the reported molar mass of the solute- too high, too low, or unaffected? Explain.
-The solute dissociating slightly will cause you to calculate a molar mass that is too small if you neglect the dissociation by assuming i=1 instead of the correct value of i, which is >1.
6. Part B.3. The temperature versus time data plot shows no change in your temperature at the freezing point for a pure solvent; however, the temperature at the freezing point for a solution steadily decreases until the solution has completely solidified. Account for this decreasing temperature.
- The freezing point of the solution is always lower than the freezing point of the pure solvent because when a non-volatile solute is dissolved in a solvent, the vapor pressure of the solvent is lowered. A depression of the solvent freezing point occurs.
7. Part C.1. Interpretation of the data plots consistently shows that the freezing points of three solutions are too high. As a result of this “misreading of the data,” will the reported molar mass of the solute be too high, too low, or unaffected? Explain.
-The change in temperature is suppose to increase because the freezing point from the cooling curves increases causing the molar mass to be lower. If three solutions are too high then the molar mass will be recorded as high.