A student dissolved 3.2 g of an unknown solute in 20.0 mL of water. The measured

Question

A student dissolved 3.2 g of an unknown solute in 20.0 mL of water. The measured freezing point was –2.1 oC . The measured freezing point of the pure solvent was -0.2 oC. Assume the density of water is 0.998 g/mL. What is the freezing point and freezing point depression of the solution? What is the molality, m, of the original solution? (m= moles solute/kg of solvent) What is the molar mass of the unknown solute? (Hint: Start by using the definition of molality to solve for moles of solute.)

can someone help me solve this?

Explanation / Answer

m = 3.2

V = 20 ml = 20 g = 0.02 Kg

Tf = -2.1

dTf = -0.2 - 2.1 = -1.9

The equation

dTf = -i*Kf*m

-1.9 = -1*1.86*m

m = 1.9/(1.86) = 1.02150 molal

m = 1.02150 molal

then

dTb = i*Kb*m = 1*0.512*1.02150 = 0.5230

Tb = 100.5230

for

1.02150 molal

kg solvnet = 0.02 then

mol = 0.02*1.02150 = 0.02043

MW = mass/mol = 3.2/0.02043 = 156.632 g /mol

MW = 156.632 g /mol

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