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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.
2. Determine the change in boiling (or freezing) point temperature (solution and pure solvent). DT=T 3. Calculate the molality, using the change in boiling point (or freezing point) and the elevation (or depression) constant. DT= mK 4. Find the moles of solute from molality by multiplying by the kg of solvent.
so 5. Now that you have the moles, plug it back into the equation from step 1 and solve for molar mass.
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; K 1. Set up this equation and place the grams on top. Now you need to find the moles to complete the problem.
2. Determine the change in boiling point temperature (solution and pure solvent). DT=T 82.42 °C -80.10 °C= 2.32 °C 3. Calculate the molality, using the change in boiling point and the elevation constant. DT= mK 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 5. Now that you have the moles, plug it back into the equation from step 1 and solve for molar mass.
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 2.07 1. Set up this equation and place the grams on top. Now you need to find the moles to complete the problem.
2. Determine the change in freezing point temperature (solution and pure solvent). DT=T 2.07 °C -5.48°C= -3.41 °C 3. Calculate the molality, using the change in boiling point and the elevation constant. DT= mK
-3.41 °C=m x (-5.12 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 5. Now that you have the moles, plug it back into the equation from step 1 and solve for molar mass.
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