Tf C for benzene is 5.5 C and Kf for benzene is 5.12 C/m Can you solve these thr

Question

Tf C for benzene is 5.5 C and Kf for benzene is 5.12 C/m

Can you solve these three problems showing all work and formulas used?

6. Freezing point depression data allow us to determine how many moles of particles are dissolved in solution. A solution of 0.5 grams of unknown and 20.0 grams of benzene freezes at 2.0°C. Use T and for benzene (see Table in lab procedure) to find the molality (mol/kg) of the solution. Show work. Use kg solvent and your calculated molality to find the moles of unknown dissolved in the 20.0 grams of solvent. Show work. Now find the molar mass of the unknown. (grams unknown / moles unknown) Show work.

Explanation / Answer

we know that

depression in freezing point = freezing point of pure solvent benzene - freezing point of solution

depression in freezing point = 5.5 - 2

depression in freezing point = 3.5 C

now

depression in freezing point = Kf x m

3.5 = 5.12 x m

m = 0.6836

so molality of the solution is 0.6836 mol/kg

2)

now

mass of solvent benzene (kg) = 0.02 kg

now

moles of solute = mass of solvent benzene (kg) x molality

moles of solute = 0.02 kg x 0.6836 mol / kg  

moles of solute = 0.01367 mol

3)

now

molar mass of solute = mass / moles

molar mass of solute = 0.5 g / 0.01367 mol

molar mass of solute = 36.57 g/mol

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