A student determines the molar mass of an unknown by the principles described ab

Question

A student determines the molar mass of an unknown by the principles described above. It was found that the freezing temperature of their pure t-butanol was 11.0 degrees Celsius on their thermometer. When they added 11.1 g of the unknown sample to the mixture the temperature fell to 2.5 degrees Celsius and the mass of the solution measured 90.4 g.

a. What was the freezing point depression?

b. What mass of t-butanol was in the solution?

c. What was the molarity of the unknown?

d. Water was the molar mass of the unknown?

e. What mass of unknown would there be in a solution containing 1kg of t-butanol at the same concentration as they had in this experiment?

Explanation / Answer

a)

dTf= Tfinal - Tinitial = 2.5-11 = -8.5°C

b)

mass of butnaol = (90.4-11.1) = 79.3 g

c)

Kf = 8.12 °C/molal

molality = dTf/Kf = 8.5/8.12 = 1.046 molal

d)

mol of unkown = molal *Kg = 1.046*79.3 /1000 = 0.0829478

Molar mass = mass/mol = 11.1/0.0829478

MM = 133.819 g/mol

e)

if m = 1 kg, then with the same concentrration -->  1.046 molal

1 kg =  1.046 moles of solute

mass = mol*MW = 1.046 *133.819 = 139.9746 g

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