Section Experiment 19 Data A. Measured B. Finding the Freezing Point of a and Ca

Question

Section Experiment 19 Data A. Measured B. Finding the Freezing Point of a and Cal Calculations: Molar Mass Determination by Depression of the Freezing Point oC Freezing Point of Pure Water g ng Point of a Solution of Liquid Unknown ters in the instructions) Target mass o Unknown # Actual mass of solute used Trial I Freezing point of solution (observed) Mass of solution Trial II Freezing point of solution Mass of solution et mass of solute (Calculated based on the parame oC 313.5o °C 2112 Calculations: Trial Trial II Co Co Freezing point depression molal molal Molality of unknown solution, m Mass of solution Mass of solute Mass of solvent (water) Moles of solute Molar mass of unknown mol mol g/mol

Explanation / Answer

Experiment 19

Solution:

B)

Calculation

Freezing Point Depression                               

T1 = Tfinal – Tinitial

      = -0.9-0.6 = -1.5

T2 = Tfinal – Tinitial

      = -0.9-0.4 = -1.3

Now, we will use T to find molality of unknown solution by assuming that i=1

Tf1 = Kfm (f = freezing) (kf = cryoscopic constant pure water: -1.86 ®C/m)

     m= Tf1/Kf

          = -1.5/-1.86 = 0.806 mol/kg = 0.8 mol/kg

Tf2 = Kfm (f = freezing) (kf = cryoscopic constant pure water: -1.86 ®C/m)

     m= Tf2/Kf

          = -1.3/-1.86 = 0.698 mol/kg = 0.7 mol/kg

Mass of solution Trial 1 = 313.50 g

Mass of solute used = 10 g

Mass of solvent (water) = 313.50 – 10 = 303.50 g

Mass of unknown/kg of water = Molar Mass * Molality

10/0.0646 = Molar mass * 0.8 mol/kg

Molar mass = 193.49 g/mol = 193.5 g/mol

Mass of solution Trial 2 = 284.92 g

Mass of solute used = 10 g

Mass of solvent (water) = 284.92 – 10 = 274.92 g

Mass of unknown/kg of water = Molar Mass * Molality

10/0.0360 = Molar mass * 0.7 mol/kg

Molar mass = 396.82 g/mol = 396.8 g/mol

C)

Freezing Point Depression

T1 = Tfinal – Tinitial

      = -0.1-0.6 = -0.7

T2 = Tfinal – Tinitial

      = -0.1-0.6 = -0.7

Now, we will use T to find molality of unknown solution by assuming that i=1

Tf1 = Kfm (f = freezing) (kf = cryoscopic constant pure water: -1.86 ®C/m)

     m= Tf1/Kf

          = -0.7/-1.86 = 0.376 mol/kg = 0.38 mol/kg

Tf2 = Kfm (f = freezing) (kf = cryoscopic constant pure water: -1.86 ®C/m)

     m= Tf2/Kf

          = --0.7/-1.86 = 0.376 mol/kg = 0.38 mol/kg

Mass of solution Trial 1 = 305.44 g

Mass of solute used = 30.533 g

Mass of solvent (water) = 305.44 – 30.533 = 274.907 g

Mass of unknown/kg of water = Molar Mass * Molality

30.533/0.274 = Molar mass * 0.38 mol/kg

Molar mass = 293.24 g/mol = 293.2 g/mol

Mass of solution Trial 2 = 312.70 g

Mass of solute used = 30.533 g

Mass of solvent (water) = 312.70 – 30.533 = 282.16 g

Mass of unknown/kg of water = Molar Mass * Molality

30.533/0.274 = Molar mass * 0.0.38 mol/kg

Molar mass = 293.24 g/mol = 293.2 g/mol

Value of i, the van’t Hoff factor

Solute needs to be identified, molar mass of fructose is 296.65 g/mol, which comes close to calculated molar mass.

I = actual molar mass/ average molar mass = 296.65/293.2 = 1.011

% ionization

Total concentration of all ions in solution = molality * i = 0.38 * 1.011 = 0.384

Molarity of solution is same as molality because we assume density is 1g/cm3.

0.38/0.384 = 98%

0.384 = (0.38 – x) +x+x

X = 0.004

0.004/0.38 = 1%

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