Question

In the Freezing Point Depression experiment, you will calculate the colligative molality of the solutions you produce, the freezing point depression, and the predicted boiling point of the resulting solutions. This question will give you practice doing this before you come to lab. In an experiment, you use 8.600 g of solvent (FP = 94.3°C; Kf = 5.2°C/m) and 0.750 g of solute (molecular mass = 163.78 g/mol). (a) How many moles of solute have you used? WebAssign will check your answer for the correct number of significant figures. mol (b) What is the colligative molality of the solution formed? WebAssign will check your answer for the correct number of significant figures. m (c) What is the predicted freezing point depression ΔTf? WebAssign will check your answer for the correct number of significant figures. °C (d) What is the predicted freezing point of the solution formed?

Answer 1

Answer:

Explanation:

(a) Moles of solute = mass/molar mass of solute

= 0.75/163.78

=0.0045793

= 0.00458 mole

(b) molality = moles of solute/mass of solvent in kg

= 0.00458/0.0086

= 0.53255 m = 0.533 m

(c) ΔTf = Kf x molality

(ΔTf = freezing point depression , Kf =molal freezing point depression)

= 5.2 x 0.533

= 2.7716 deg C = 2.8 deg C

(d) Freezing point of solution = freezing point of solvent - ΔTf

= 94.3 - 2.8

= 91.5 deg C