What would be the effect of each of the following on the calculated molecular weight of the solute?
a) some cyclohexane evaporated while the freezing point of pure cyclohexane was being measured. ( i think it would be higher)
b)Some cyclohexane evaporated after the solute was added. ( i think it would be lower)
c) a foreign solute was already present in the cyclohexane. (i think it would have no change)
d) the thermometer is not calibrated correctly. it give a temp that is 1.5 C too low at all temperatures. (i think it would be higher)
I'm going to differ with both of you.Here are the formulas used to solve for molecular weight.
(1)...delta T = Kf*m
(2)...m = mols/kg solvent
(3)...mol weight = grams/mols
a. no effect. The normal freezing point of pure cyclohexane will be same whether you determine the freezing point on 1 mL or on 100 mL.
b. I think lower is right. Why? If some of the solvent evaporated after adding the solute, then m is higher from 2) which makes mols higher. mols higher in 3) makes molecular weight smaller.
c. I think c is right at no change. The foreign material is constant and the effects are constant so delta T will be the same (although T initial and T final may not be the same).
d. no change. So the thermometer reads 1.5 too low every time it's read. T final and T initial readings will not be right BUT the difference (delta T) will always be the same and that's all you're using in the calculations.
Can not be positive, but switch your answers for c and d, and your answer for a should say lower.
Didn't even think to look at it that way.
Actually, it is not pure cyclohexane, cyclohexane is the solvent in the solution.That should have been specified to get an accurate response.
To determine the effect on the calculated molecular weight of the solute, we need to understand the principles behind the freezing point depression method.
In the freezing point depression method, the molecular weight of a solute can be calculated by measuring the difference between the freezing point of the pure solvent and the freezing point of the solution. By using the formula:
ΔT = Kf * m,
where ΔT is the change in freezing point, Kf is the cryoscopic constant of the solvent, and m is the molality of the solution, we can determine the molecular weight of the solute.
Now, let's analyze each scenario individually:
a) If some cyclohexane evaporated while measuring the freezing point of pure cyclohexane, the observed freezing point of the pure solvent would be lower than expected. This is because evaporation removes the more volatile component of the solution, leaving behind a higher concentration of the solute. Consequently, the calculated molecular weight of the solute would be higher.
b) If some cyclohexane evaporated after the solute was added, the observed freezing point of the solution would be higher than expected. This is because the evaporation of the solvent increases the concentration of the solute in the remaining solution, resulting in a greater freezing point depression. Consequently, the calculated molecular weight of the solute would be lower.
c) If a foreign solute was already present in the cyclohexane, the presence of the foreign solute would affect the freezing point of the solution. The molecular weight of the solute being measured would not be directly affected in this case. However, it is worth noting that the foreign solute could interfere with the measurement and potentially affect the accuracy of the calculated molecular weight.
d) If the thermometer is not calibrated correctly and consistently measures temperatures 1.5 degrees Celsius too low, the measured freezing point of the solvent or solution would be lower than the actual freezing point. This would lead to an overestimation of the freezing point depression and hence an overestimation of the molecular weight of the solute.
In summary:
a) Evaporation of cyclohexane during the measurement of freezing point depression would result in a higher calculated molecular weight for the solute.
b) Evaporation of cyclohexane after the solute was added would result in a lower calculated molecular weight for the solute.
c) The presence of a foreign solute does not directly affect the calculated molecular weight of the solute but may interfere with the measurement.
d) A thermometer that consistently reads 1.5 degrees Celsius too low would result in a higher calculated molecular weight for the solute.