1) What would be the effect of each of the following on the calculated molecular
ID: 1003826 • Letter: 1
Question
1) What would be the effect of each of the following on the calculated molecular weight of a solute?a) Some cyclohexane evaporated while the freezing point of pure cyclohexane was being measured.
b). Some cyclohexane evaporated after the solute was added.
c). A foreign solute was already present in the cyclohexane.
d). The thermometer is not calibrated correctly. It gives a temperature that is 1.5 degrees Celcius too low at all temperatures. 1) What would be the effect of each of the following on the calculated molecular weight of a solute?
a) Some cyclohexane evaporated while the freezing point of pure cyclohexane was being measured.
b). Some cyclohexane evaporated after the solute was added.
c). A foreign solute was already present in the cyclohexane.
d). The thermometer is not calibrated correctly. It gives a temperature that is 1.5 degrees Celcius too low at all temperatures.
a) Some cyclohexane evaporated while the freezing point of pure cyclohexane was being measured.
b). Some cyclohexane evaporated after the solute was added.
c). A foreign solute was already present in the cyclohexane.
d). The thermometer is not calibrated correctly. It gives a temperature that is 1.5 degrees Celcius too low at all temperatures.
Explanation / Answer
Tf = Kf.m ;m = molality of the solution = no of moles of the solute per kg of the solvent
and no of moles = mass/molar mass
1. a. freezing point of pure solvent is not effected by evaporation of the solvent (cyclohexane). So this will not effect the calculated molar mass of the solvent.
b. no of moles = mass/molar mass
When some portion of the solvent evaporates, concentration(i.e., molality, m) of the solution increases as mass of the solute remains unchanged. As a result the freezing point of the solution
will be lower than it should be. The calculated molar mass of the solvent will be smaller than it should be as the no of moles in the solution increases but the mass of the solute remains unchanged.
c. If a foreign solute is present in the solution, it will effect the freezing point of both the pure solvent and solution. As the deviation is same for both cases, Tf will not change and this will not effect the calculated molar mass of the solvent.
d. No effect, as Tf will remain same because the deviation is same for both cases- pure solvent and the solution.
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