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.
Please someone explain this to me I don't understand it...Thank you
Are you measuring the f.p. of the cyclohexane solvent or the freezing point of the solute? If the solute, what is the solute? Are you using the depression of freezing point method?
freezing point of cyclohexane...yes I am using the depression of freezing point method
If there is no solute in the cyclohexane then the amount will not affect the f.p. In other words, the f.p. of cyclohexane is the same whether you've tested 1 mL or 100 mL
The loss of cyclohexane vapor during the temperature versus time measurement will result in the recorded freezing point being too low. This is because when a liquid evaporates, it takes energy from its surroundings. In the case of cyclohexane, as it evaporates, it absorbs heat from the surroundings, which causes the remaining liquid to cool down.
In a typical temperature versus time measurement for a substance undergoing phase transition, the temperature is expected to remain constant during the phase change. This is because the absorbed heat is being used to break the intermolecular bonds and convert the substance from a liquid to a solid state (freezing). However, with the loss of cyclohexane vapor, the remaining liquid will have less energy and therefore a lower temperature.
As a result, the recorded freezing point will be too low because the temperature will not be constant during the phase change. The actual freezing point of the cyclohexane would be higher, but due to the loss of solvent vapor, the observed freezing point will be lower than expected.