Question

You obtained the following graphical data from two experimental runs to determine the freezing point depression constant for a pure solvent. The solution was made up by adding 3.025 g of naphthalene, C₁₀H₈, to 45.320 g of pure solvent.

85.5 Plot for determining freezing point depression constant 84.5 83.5 Temperature, 82.5 Solution cooling curve 81.5 Solvent cooling curve 80.5 79.5 0 100 200 300 400 500 600 Time, s

1.What is the freezing point temperature of the pure solvent (in ^oC)?

2.What is the solution freezing point temperature (in ^oC)?

3.What is the freezing point depression, T_fp (in ^oC)?

4.What is the molar mass of naphthalene (in g/mol)?

5.How many moles of naphthalene are in the solution?

6.What is the molality of the solution (in mol/kg-solvent)?

7.What is the freezing point depression constant, k_fp, of the pure solvent (in ^oC/m)?

Answer 1

1) The solvent temperature is 82.5 ° C

2) The temperature of the solution is 79.6 ° C.

3) The depression is calculated:

ΔT = Ts - Tsl = 82.5 - 79.6 = 2.9 ° C

4) The molar mass of naphthalene is 128.2 g / mol.

5) The moles of naphthalene are calculated:

n = m / MM = 3,025 g / 128.2 g / mol = 0.024 mol

6) The molality is calculated:

m = n / Kg = 0.024 mol / 0.04532 Kg = 0.530 m

7) The constant is calculated:

Kf = ΔT / m = 2.9 ° C / 0.53 m = 5.47 ° C / m

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