Help me solve this

Question

You are studying an unknown substance that appears to be some type of sugar. You prepare a solution by dissolving 24.48g of the sugar in 160.0mL of water and find that the freezing point of the resulting solution is -1.153°C. What is the molecular mass of this unknown sugar? The freezing point depression constant for water is 1.86°C/molal. Assume the sugar is a molecular solute and report your answer in units of "amu"

Answer 1

∆T = imK

∆T = change in freezing point = 1.153ºC

i = van't Hoff factor = 1 for a non electrolyte (sugar is a non electrolyte)

m = molality = mols solute / kg solvent = ?

K = freezing constant = 1.86º/m

Solving for m, we have ...

m = ∆T/(i)(K) = 1.153ºC / (1)(1.86º/m)

m = 0.61989 mols/kg

Since we have 160.0 ml of solvent, and assuming a density of 1 g/ml, we have 160.0 g solvent.

This is equal to 0.1600 kg solvent

Thus ... 0.160 kg x 0.61989 mols/kg = 0.09918 mols

Molar mass = g / moles

molar mass = 24.48 g / 0.09918 mols

molar mass = 246.8 g / mol = 246.8 amu = 247 amu (3 sig.figs.)

Answer 2

This is a question about freezing point depression, which is when dissolving a nonvolatile solute results in lowering the freezing point (or raising the boiling point). This is what we call a colligative property, meaning it depends only on the concentration of the solute particles, and not on the solute's identity.

The relevant equation is:

delta T = i * Kf * M

Here, M is the molality of the solution (concentration in moles solute per kg solvent), Kf is the freezing point depression constant (a property of the solvent, which they gave us for water), and i is the Van't Hoff factor, and equals the number of particles/ions the solute dissociates into.

Sugar is molecular rather than ionic, and is not a strong acid or base, so it shouldn't dissociate at all when it is dissolved. Since it stays all in one piece, the Van't Hoff factor is one.

We are given delta T and Kf, so we can go ahead and plug all of this in and solve for the molality.

1.153 degrees C = 1 * 1.86 degrees C per molal * M

M = 1.153/1.86 molal = 0.6199 molal

Remember, molality is moles of solute per kg of solvent.

0.6199 molal = 0.6199 moles sugar / 1 kg water.

So we can use this, together with how much water was used to make the solution, to determine the number of moles of solute. The density of water is 1 g/mL.

160.0 mL * (1g/mL) * (1 kg/1000 g) * (0.6199 moles/kg) = 0.09918 moles.

To get the molar mass in g/mol, which will be the same as the molecular mass in amu, we just take the grams of solute needed to make the solution by the moles we just calculated.

24.48 g / 0.09918 moles = 246.8 g/mol = 246.8 amu

But how many sig figs? Most of the numbers we're provided have 4 sig figs, but the freezing point depression constant only has 3, so we'll use 3 in our answer. (We use the rules for multiplication and division.)

molecular mass = 247 amu