Question

The freezing point of benzene, C6H6, is 5.500 °C at 1 atmosphere. Kf(benzene) = 5.12 °C/m In a laboratory experiment, students synthesized a new compound and found that when 14.44 grams of the compound were dissolved in 292.3 grams of benzene, the solution began to freeze at 4.571 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound?

Answer 1

Answer:

The molecular weight they determined for this compound is 272.9 g/m

Explanation:

We must apply the colligative property of freezing point depression to solve this:

ΔT° = Kf · molality

Where ΔT° means T° fussion pure solvent − T° fussion of solution

and Kf the cryoscopic constant

The statemente expressed that the compound was also found to be nonvolatile and a non-electrolyte, so we don't have to apply the Van't Hoff factor (i)

By the way, the complete formula is this one:

ΔT° = Kf · molality . i

5.5°C - 4.571°C = 5.12 °C/molal . molality

0.929°C = 5.12 °C/molal . molality

0.929°C / 5.12 molal/°C = molality → 0.181 m

Molality means the mol of solute in 1kg of solvent. But, in this solution we used 292.3 g of benzene, so let's find out our moles of solute, in our mass of solvent.

1000 g ___ 0.181 moles

292.3 g ____ (292.3 . 0.181) /1000 = 0.053 moles

The mass, we used of solute is 14.44 g so, to find out the molar mass we must divide mass (g) / moles

14.44 g /0.053 m = 272.9 g/m