MASINDE MULIRO UNIVERSITY OF SCIENCE AND TECHNOLOGY

SCHOOL OF NATURAL SCIENCES
Department of Pure and Applied Chemistry
SCH 140: Physical Chemistry
CAT 2 18/01/2021
Instructions:
Each question has 5 marks
Due Date: Friday, 22nd January 2021 latest 5 pm
Submission through respective class reps

1. A crystalline white solid is a mixture of glucose (C6H12O6) and sucrose
(C12H22O11). Is it possible that a 10.00g sample of the solid dissolved in 100.0 g
H2O might have a freezing point of -1.25oC? Explain.

Assume two scenarios; i.e., the 10 g sample is 100% glucose or is 100% sucrose.

A. If 100% glucose then
mols glucose = grams/molar mass = 10.00/180 = 0.0555
molality = mols/kg solvent = 0.0555/0.100 = 0.555
dT = i*kf*m = 1*1.86*0.555 = 1.03 so freezing point is -1.03 so glucose won't do it.
B. If 100% sucrose. Since molar mass sucrose is higher than that of glucose (342 vs 180) then mols sucrose will be lower, molality will be lower, delta T will be lower and freezing point will be higher. You can confirm all of this--I didn't want to do the calculation.
So the answer to the problem is no, the freezing point cannot be as low as -1.25 C.

cocaine

To determine if it is possible for a 10.00g sample of the crystalline white solid (mixture of glucose and sucrose) dissolved in 100.0g of water (H2O) to have a freezing point of -1.25°C, we need to consider the colligative property of freezing point depression.

The freezing point depression is a phenomenon that occurs when a solute is dissolved in a solvent, causing the freezing point of the solution to be lower than that of the pure solvent. This lowering of the freezing point is directly proportional to the concentration or molality (moles of solute per kilogram of solvent) of the solute particles.

In this case, glucose (C6H12O6) and sucrose (C12H22O11) are both non-electrolyte solutes, meaning they dissociate fully into individual solute particles when dissolved in water. Since they do not produce ions in solution, they do not affect the colligative properties as strongly as ionic solutes.

To calculate the freezing point depression, we can use the following formula:

ΔT = Kf * m

Where:
ΔT is the change in freezing point (in degrees Celsius),
Kf is the cryoscopic constant (a property of the solvent, in this case water),
m is the molality of the solute in the solution.

The cryoscopic constant for water (Kf) is 1.86 °C/m.

First, we need to calculate the molality (m) of the solution. Molality is defined as the moles of solute per kilogram of solvent.

For glucose:
Molecular weight (C6H12O6) = 180.16 g/mol

Number of moles of glucose = mass of glucose / molar mass of glucose
= 10.00g / 180.16 g/mol

Molality of glucose (m) = moles of glucose / mass of water (in kg)
= (10.00g / 180.16 g/mol) / (100.0g / 1000)
= (10.00g / 180.16 g/mol) / 0.1 kg

Similarly, we can calculate the molality of sucrose using its molecular weight (C12H22O11).

Once we have the molality of the solute in the solution, we can calculate the change in freezing point (ΔT) using the formula above.

Finally, we compare the calculated change in freezing point (ΔT) with the given value of -1.25oC. If the calculated value matches the given value, then it is possible for the solution to have a freezing point of -1.25°C. Otherwise, it is not possible.

Note: The actual calculation of ΔT and comparison with the given value should be done using the specific values obtained for molality and cryoscopic constant.