Asucrose solution is made using 144 g of sucrose (C12H22O11) and diluting to a total volume of 1.00 L. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.06 g/mL for the final solution.)

Solution 122P:

Here, we are going to calculate the boiling point and freezing point of the solution .

Step 1: Calculation of the amount of sucrose in mol

We know,

Molar mass of sucrose =342.296 g/mol

Mass of sucrose = 144 g

Amount of glucose in mol = 144g/342.296 g/mol = 0.4207 mol

Step 2: Calculation of molality of the solution.

Density = Mass/volume = 1.06 g/mL

Volume = 1L= 1000 mL

Mass of solution = Density x volume = 1.06 g/mL x 1000 mL = 1060 g

Volume of solution = 1060g = 1.06 kg

Molality (m) of the solution = 0.4207 mol/1.06 kg =0.3968 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

= 0.3968 m x 1.86 oC/m= 0.7382 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.

Thus,

Tf = T∘f - ΔTf = 0 - 0.7382 oC = -0.7382 oC

Therefore , the freezing point of the solution is -0.7382 oC or -0.74 oC