Posted by Jim on .
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)
Can not be positive, but switch your answers for c and d, and your answer for a should say lower.
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.
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.