Question

What is the effect on your calculated molar mass if the following happened:

a. The team lost some t-butanol solvent prior to adding the solid unknown and starting the determination of the freezing point of the solute/solvent mix but kept the original t-butanol mass in their calculations.

b. The team massed out 0.9000g of unknown solute but only got 0.7000g into the t-butanol solvent but used the 0.9000g in their calculations.

Answer 1

1. Calculation of freezing point as

∆Tf = i. kf. m

Where i = vant hoff’s factor

Kf = freezing point depression constant

M = molality

Hence freezing point depends on the molality

Molality m = no. of moles of solute/mass of solvent in kg

Hence freezing point depression depends on the mass of the solvent.

The team lost some mass of the solvent in beginning during experiment but while doing the calculations they kept the original mass of the solvent.

As correct value of the mass of solvent is used, in the calculated Molar mass will be correct.

No effect.

b. As correct mass of the solvent is used while doing the calculations, hence it will not effect the calculation for molar mass of unknown.

correct mass of solvent 0.9000g is used so, result will be correct.