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
