Experiment: Molar mass determination by freezing point depression in 2-methyl-2-

Question

Experiment: Molar mass determination by freezing point depression in 2-methyl-2-propanol. Solvent is 2-methyl-2-propanol, solute is unknown.

How would each of the following errors change the value of the molecular mass of the solute obtained in this experiment? Compare with the true value of molecular mass. (Assume true value is 100.0 g/mol). Choose one: less, the same, greater. Explain your reasoning.

a) Some solvent is lost through the experiment.

b) The compound (solute) was not completely dissolved.

c) A foreign solute was already present in the solvent.

d) The thermometer is not calibrated correctly and as a result it displays temperatures that are always 1.2°C higher than the actual temperature

Explanation / Answer

a) If some solvent is lost though experiment then the mass of solvent will be less as compared to expected

Molality = Moles of solute / Mass of solvent

Molality = Mass / Molecular weight x mass of solvent

Depression in freezing point = Kf X molality

So if mass is less the molality will be more as expected, so the depression in freezing point will be more and hence the molecular weight will be low (Less)

b) if the compound was not completely dissolved then less mass will be dissolved in solvent less depression in freezing point hence more molecular weight

c) if a foreign solute is already present then that is also present thought out the experiment so the depression in freezing point will be the same and hence the molecular weight will be unchanged

d) if therometer is not calibrated correctly then it will read with same error initially and finally as well after the addition of solute

so the change in temperature (which is measured as depression in freezing point) will be the same.

So the molecular weight will be the same

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