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

Answer: No. The freezing point would be -1.03 degrees C

I don't think this problem can be worked exactly since there is no indication of how the mixture is divided between the two sugars. You can, however, calculate the freezing point assuming it is pure glucose and assuming it is pure sucrose. The 1.03 cited in the answer is assuming pure glucose.

mols glucose = grams/molar mass
m = mols/kg solvent
delta T = Kf*m = about =-1.03

Do the same for sucrose.
I get about -0.5.
So no matter the composition, it can't make it to -1.25.

To determine if a 10.00 g sample of the solid dissolved in 100.0 g H2O can have a freezing point of -1.25 degrees C, we need to consider the properties and behavior of the solute (glucose and sucrose) and the solvent (water) in a solution.

The freezing point depression is a colligative property, which means it depends on the number of solute particles in a given amount of solvent. The more solute particles present, the greater the freezing point depression.

In this case, the solute consists of both glucose (C6H12O6) and sucrose (C12H22O11), both of which are organic compounds with different molecular formulas and molar masses.

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

ΔT = Kf * m * i

ΔT: Freezing point depression
Kf: Freezing point depression constant for the solvent
m: Molality, which is the moles of solute per kilogram of solvent
i: Van't Hoff factor, which represents the number of particles the solute dissociates or associates into in the solution.

In this case, we have a solution with a mass of 100.0 g H2O (solvent) and 10.00 g of the solid (solute). We need to find the number of moles of each solute to calculate both the molality and the Van't Hoff factor.

Molar mass of glucose (C6H12O6) = 180.16 g/mol
Molar mass of sucrose (C12H22O11) = 342.30 g/mol

For glucose:
moles of glucose = mass / molar mass
moles of glucose = 10.00 g / 180.16 g/mol = 0.0555 mol

For sucrose:
moles of sucrose = mass / molar mass
moles of sucrose = 10.00 g / 342.30 g/mol = 0.0292 mol

Now, let's calculate the molality (moles of solute per kilogram of solvent):

molality = moles of solute / mass of solvent (in kg)
mass of solvent = 100.0 g H2O = 0.1000 kg

For glucose:
molality of glucose = 0.0555 mol / 0.1000 kg = 0.555 mol/kg

For sucrose:
molality of sucrose = 0.0292 mol / 0.1000 kg = 0.292 mol/kg

The Van't Hoff factor will depend on whether the solute particles in the solid dissociate or associate in the solution. In this case, glucose does not dissociate or associate, so its Van't Hoff factor (i) is 1. On the other hand, sucrose dissociates into glucose and fructose in the solution, so its Van't Hoff factor (i) is 3.

Now, let's calculate the freezing point depression using the given freezing point constant for water (Kf = 1.86 °C/m):

ΔT = Kf * m * i

For glucose:
ΔTglucose = 1.86 °C/m * 0.555 mol/kg * 1 = 1.03 °C

For sucrose:
ΔTsucrose = 1.86 °C/m * 0.292 mol/kg * 3 = 1.62 °C

Since the freezing point depression is a colligative property, we need to consider the total moles of solute particles present in the solution. Therefore, we need to sum up the contributions from both glucose and sucrose:

Total ΔT = ΔTglucose + ΔTsucrose
Total ΔT = 1.03 °C + 1.62 °C = 2.65 °C

Therefore, it is not possible for a 10.00 g sample of the solid dissolved in 100.0 g H2O to have a freezing point of -1.25 degrees C. The freezing point depression would be approximately -2.65 degrees C.