Data and Caloulationsi Molar Mass Determination by Depression of the Freezing Po
ID: 564407 • Letter: D
Question
Data and Caloulationsi Molar Mass Determination by Depression of the Freezing Point A Measured Freezing Point of Pure Water B. Finding the Reezing Point of a Solution ot uquld Unknown Turget mass of solute (Calculated based on the parameters in the in structions Unknown e Actual mass of solute used Trial I Freezing point of solution (observed) Mass of solution Trial II Freezing point of solution Mass of solution "c 103. 81 -5, 4 Calculations: Trial Trial 11 5. a Freezing point depression Molality of unknown solution, m, Mass of solution Mass of solute Mass of solvent (water) Moles of solute Molar mass of unknownExplanation / Answer
Assuming the solute does not dissociate.
For trial 1
Freezing point depression
Tf - Ts = i x Kf x m
Freezing point of water - Freezing point of solution = van't Hoff factor x freezing point constant x Molality
-0.2 + 4.6 = 1 x 1.86 x m
Molality of unknown solution = 2.365 molal
Mass of solution = 103.81 g
Mass of solute = 10.40 g
Mass of water = Mass of solution - Mass of solute
= 103.81 - 10.40
= 93.41 g
Moles of solute = molality x kg of water
= 2.365 x 93.41 g x 1kg/1000g
= 0.2209 moles
Molar mass of unknown = mass of solute / moles of solute
= 10.40 g / 0.2209 moles
= 47.08 g/mol
Same for trial 2
Freezing point depression
Tf - Ts = i x Kf x m
Freezing point of water - Freezing point of solution = van't Hoff factor x freezing point constant x Molality
-0.2 + 5.4 = 1 x 1.86 x m
Molality of unknown solution = 2.7956 molal
Mass of solution = 95.08 g
Mass of solute = 10.40 g
Mass of water = Mass of solution - Mass of solute
= 95.08 - 10.40
= 84.68 g
Moles of solute = molality x kg of water
= 2.7956 x 84.68 g x 1kg/1000g
= 0.2367 moles
Molar mass of unknown = mass of solute / moles of solute
= 10.40 g / 0.2367 moles
= 43.937 g/mol
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