EPRESSION - Colligative properties of a solution are properties that depend on the number of solute particles dissolved. - Colligative properties of a solution include: o Vapor pressure lowering, o Boiling point elevation, o Freezing point depression, and o Osmotic pressure. • The molar mass of a solute may be determined by measuring a colligative property of a solution. • In this experiment the freezing point of a solvent would have been determined from a cooling curve. actice problems: • Calculate the freezing point of a solution of 800 g of ethylene glycole (MW = 62.02 g/mole) in 3,000 g of water given that the K(H2O) = 1.86 °C/m. • How many moles of ethanol are dissolved in t-butanol if the change in the freezing point is 10 °C, and t-butanol's Kr is 8.37 °C/m? How many grams of ethanol are dissolved in this solution? Calculate the freezing point of a solution of 200 g of glucose (MW = 180.096 g/mole) in 2,500 g of acetic acid, given that K(acetic acid) = 3.90 °C/m, and T (acetic acid) = 16.6 °C. • Calculate the boiling point of a solution of 280 g of glucose (MW = 180.096 g/mole) in 3,600 g of acetic acid, given that K»(acetic acid) = 3.07 °C/m, and T'b(acetic acid) = 118.5 °C. • Calculate the boiling point of a solution of 15 g of KBr dissolved in 350 g of water, given that KB(H2O) = 0.52 °C/m.

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Chapter80: Crystallization: Purification Of Solids
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LAB VI. MOLAR MASS BY FREEZING POINT DEPRESSION
• Colligative properties of a solution are properties that depend on the number of solute
particles dissolved.
• Colligative properties of a solution include:
o Vapor pressure lowering,
o Boiling point elevation,
o Freezing point depression, and
o Osmotic pressure.
• The molar mass of a solute may be determined by measuring a colligative property of a
solution.
• In this experiment the freezing point of a solvent would have been determined from a
cooling curve.
Practice problems:
• Calculate the freezing point of a solution of 800 g of ethylene glycole (MW = 62.02
g/mole) in 3,000 g of water given that the K(H2O) = 1.86 °C/m.
• How many moles of ethanol are dissolved in t-butanol if the change in the freezing
point is 10 °C, and t-butanol's Kr is 8.37 °C/m? How many grams of ethanol are
dissolved in this solution?
Calculate the freezing point of a solution of 200 g of glucose (MW = 180.096 g/mole)
in 2,500 g of acetic acid, given that Kr(acetic acid) = 3.90 °C/m, and T(acetic acid)
= 16.6 °C.
Calculate the boiling point of a solution of 280 g of glucose (MW = 180.096 g/mole)
in 3,600 g of acetic acid, given that Kb(acetic acid) = 3.07 °C/m, and T(acetic acid)
= 118.5 °C.
Calculate the boiling point of a solution of 15 g of KBr dissolved in 350 g of water,
given that K(H2O) = 0.52 °C/m.
Transcribed Image Text:LAB VI. MOLAR MASS BY FREEZING POINT DEPRESSION • Colligative properties of a solution are properties that depend on the number of solute particles dissolved. • Colligative properties of a solution include: o Vapor pressure lowering, o Boiling point elevation, o Freezing point depression, and o Osmotic pressure. • The molar mass of a solute may be determined by measuring a colligative property of a solution. • In this experiment the freezing point of a solvent would have been determined from a cooling curve. Practice problems: • Calculate the freezing point of a solution of 800 g of ethylene glycole (MW = 62.02 g/mole) in 3,000 g of water given that the K(H2O) = 1.86 °C/m. • How many moles of ethanol are dissolved in t-butanol if the change in the freezing point is 10 °C, and t-butanol's Kr is 8.37 °C/m? How many grams of ethanol are dissolved in this solution? Calculate the freezing point of a solution of 200 g of glucose (MW = 180.096 g/mole) in 2,500 g of acetic acid, given that Kr(acetic acid) = 3.90 °C/m, and T(acetic acid) = 16.6 °C. Calculate the boiling point of a solution of 280 g of glucose (MW = 180.096 g/mole) in 3,600 g of acetic acid, given that Kb(acetic acid) = 3.07 °C/m, and T(acetic acid) = 118.5 °C. Calculate the boiling point of a solution of 15 g of KBr dissolved in 350 g of water, given that K(H2O) = 0.52 °C/m.
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