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 10.69 grams of the compound were dissolved in 260.5 grams of benzene, the solution began to freeze at 4.957 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound? _____g/mol

Answer 1

delta Tf = Kf * m

where, m = molality

delta Tf = freezing point of pure solvent - freezing point of solution

delta Tf = 5.500 - 4.957 = 0.543 ⁰C

delta Tf = Kf * molality

0.543 = 5.12 * molality

molality = 0.543 / 5.12 = 0.106 m

molality = number of moles of solute / mass of solution in kg

0.106 mole = number of moles of solute / 0.2605 kg

number of moles of solute = 0.106 * 0.2605 = 0.0276 mole

number of moles of solute = mass of solute / molar mass of solute

0.0276 mole = 10.69 g / molar mass of solute

molar mass of solute = 10.69 g / 0.0276 mole = 387 g/mol

Therefore, the molecular weight of compound = 387 g/mol