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 Molar Mass from Boiling Point Elevation or Freezing Point Depression

The addition of a nonvolatile solute to a solvent causes the boiling point of the solvent to increase and the freezing point of the solvent to decrease.

molar mass= grams /moles  , so we need to find the grams and divide that # by the number of moles

1.  Set up this equation and place the grams on top. Now you need to find the moles to complete the problem.

 molar mass= grams moles

2.  Determine the change in boiling (or freezing) point temperature (solution and pure solvent).

DT=Tsoln -Tpure

3. Calculate the molality, using the change in boiling point (or freezing point) and the elevation (or depression)  constant.

DT= mKb    or   DT= mKf

4. Find the moles of solute from molality by multiplying by the kg of solvent.

 molality (m)= moles solute kg of solvent

so

molality (m)  x kg of solvent = moles of solute

5. Now that you have the moles, plug it back into the equation from step 1 and solve for molar mass.

 molar mass= grams moles

Example 1   Molar Mass from Boiling Point Elevation

5.00 g of an organic solid is dissolved in 100.0 g of benzene. The boiling temperature of this solution is 82.42 °C. The boiling temperature of pure benzene is 80.10 °C; Kb = 2.53 °C /m. What is the molecular weight of the unknown compound?

1.  Set up this equation and place the grams on top. Now you need to find the moles to complete the problem.

 molar mass= 5.00 grams X  moles

2.  Determine the change in boiling  point temperature (solution and pure solvent).

DT=Tsoln -Tpure

82.42 °C -80.10 °C= 2.32 °C

3. Calculate the molality, using the change in boiling point  and the elevation constant.

DT= mKb

2.32 °C=m x 2.53°C /m

m= 0.917m = 0.917mol/kg

4. Find the moles of solute from molality by multiplying by the kg of solvent.

100.0g benzene x 1kg/1000g= 0.1000kg

0.917mol/kg x 0.1000kg = 0.0917 moles of solute

5. Now that you have the moles, plug it back into the equation from step 1 and solve for molar mass.

 molar mass= 5.00 grams = 54.5g/mol 0.0917 moles

Example 2  Molar Mass from Freezing Point Depression

The freezing point of a solution that contains 1.00 g of an unknown compound, dissolved in 10.0 g of benzene is found to be 2.07oC. The freezing point of pure benzene is 5.48 oC. The molal freezing point depression constant of benzene is 5.12oC /m. What is the molecular weight of the unknown compound?

1.  Set up this equation and place the grams on top. Now you need to find the moles to complete the problem.

 molar mass= 5.00 grams X  moles

2.  Determine the change in freezing point temperature (solution and pure solvent).

DT=Tsoln -Tpure

2.07 °C -5.48°C= -3.41 °C

3. Calculate the molality, using the change in boiling point  and the elevation constant.

DT= mKf

-3.41 °C=m x (-5.12oC /m)

m= 0.666m = 0.666 mol/kg

4. Find the moles of solute from molality by multiplying by the kg of solvent.

10.0g benzene x 1kg/1000g= 0.0100kg benzene

0.666 mol/kg x 0.0100kg = 0.00666 moles of solute

5. Now that you have the moles, plug it back into the equation from step 1 and solve for molar mass.

 molar mass= 1.00 grams = 150.g/mol 0.00666 moles

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