A 250.0-mL sample of a 5.00 M glucose (C6H12O6) solution is diluted to 1.40 L. What are the freezing and boiling points of the final solution? (Assume a density of 1.06 g/mL for the final solution.)

Problem 123P

A 250.0-mL sample of a 5.00 M glucose (C6H12O6) solution is diluted to 1.40 L. What are the freezing and boiling points of the final solution? (Assume a density of 1.06 g/mL for the final solution.)

Solution 123P:

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

Step 1: Calculation of the strength of the final solution by using dilution equation.

M1V1= M2V2 ------(1)

Where, M1 and V1 are Molarity and volume of the initial solution whereas M2 and V2 are the Molarity and volume of the solution.

From the Question,

It is given that,

M1 (initial Molarity) =5 M

V1 (initial volume) = 250.0 mL

M2 (final Molarity) =?

V2 (initial volume) = 1.40L =1400 mL

Thus, from equation (1)

M1V1= M2V2 ------(1)

M2 = M1V1/V2

= = 0.8928 M

The final strength of the solution is 0.8928 M = 0.8928 mol/L

1 L solution contains = 0.8928 mol

1.40 L solution contains = 0.8928 mol x 1.40L/1 L = 1.25 mol sucrose

Step 2: Calculation of the amount of glucose in mol

We know,

Molar mass of sucrose =180.156 g/mol

The final strength of the solution is 0.8928 M = 0.8928 mol/L

1 L solution contains = 0.8928 mol

1.40 L solution contains = 0.8928 mol x 1.40L/1 L = 1.25 mol sucrose

Step 2: Calculation of molality of the solution.

Density = Mass/volume = 1.06 g/mL

Volume = 1.40L= 1400 mL

Mass of solution = Density x volume = 1.06 g/mL x 1400 mL = 1484 g

Volume of solution = 1484 g = 1.48 kg

Molality (m) of the solution = 1.25 mol/1.48 kg =0.84459 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

= 0.84459 m x 1.86 oC/m= 1.570 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 - 1.570 oC = -1.57 oC

Therefore , the freezing point of the solution is -1.57 oC