Question

What would be the effect of each of the following on the calculated molar mass of the solute?

1. some cyclohexane evaporated while the freezing point of pure cyclohexane was being measured.

2. some cyclohexane evaporated after the solute was added.

3. some unknown power was still left one the weighing paper.

4. a foreign solute was already present in the cyclohexane?

5. the thermometer is not calibrated correctly. it gives a temperature that is 1.5 C too low at all temperatures.

6. solution didn't freeze completely.

please help me answer these. thank you! i promise to rate you :)

Answer 1

molcular mass of solute (M) = mass in grams / molality ---------------1

ΔT_f  = mXK_f -------------2 Kf  = cryoscopic constant ;   ΔT_f  = depression in freezing point; m= molality

from equation 2 it can be concluded that any factor that increases ΔT_f will increase in molality (m). when molality increases molecular mass will decrease. (According to 1 inverse relation between M and m).

any factor that increases the W2 will lead to increased molecular mass.

1) and 2). In 1 the freezing point of pure cyclohexane will not change but that of solution will decrease. Because as amount of solvent decreases by evaporation the molality will increase. Depression in freezing point will also be enhanced. As molality increase Molar mass (M)) will decrease. Same will happen in 2.  

3) Due to the presence of unknown powder on weighing paper weight of measured solute will be less hence molarity will also be less thn expected. Hence M will also be less.

4) There will be more depression in freezing point (ΔT_f will be higher) hence molality will be higher. So molecular mass will be less. r.

5) Ther will be no effect on  ΔT_f hence on M2.  ΔT_f = T_solution - T_solvent. Both T_solution and T_solvent will be less by 1.5C hnce no over all effect on ΔT_f .

6) if solution does not freeze well T_solution will be less hence ΔT_f will alsobe lesser . therefore molality and hence M will be higher.