A glucose solution contains 55.8 g of glucose (C6H12O6) in 455 g of water. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)
Solution 105 P:
Here, we are going to calculate the boiling point and freezing point of the solution .
Step 1: Calculation of the amount of glucose in mol
Molar mass of glucose = 180.15 g/mol
Mass of glucose = 55.8 g
Amount of glucose in mol = 55.8 g/180.1559 g/mol = 0.3097 mol
Step 2: Calculation of molality of the solution.
Volume of solution = 455 g =0.455 kg
Molality (m) of the solution = 0.3097 mol/0.455 kg = 0.6807 m
Step 3: Calculation of freezing point of the solution.
The freezing point of depression of a solution is calculated by the equation:
Tf= m x Kf
Where m is the molality of the solution and Kf is the freezing point of depression constant for the solvent.
Tf= m x Kf , where kf = 1.86 oC/m
= 0.6807 m x 1.86 oC/m= 1.266 oC
The freezing-point depression can be calculated as
ΔTf=T∘f−Tf , where T∘f - the freezing point of the pure solvent (water) = 0 oC
Tf - the freezing point of the solution.
Tf = T∘f - ΔTf = 0 - 1.266 oC = -1.266 oC = -1.27oC
Therefore , the freezing point of the solution is -1.27oC