I have the answers, I am just interested in the formulas so that I can solve the problems, thanks!

1. Find the boiling point of a solution composed of 110g of HgCl2 (a non-ionizing solute) in 175g of water.

2. Find the freezing point of a solution composed of 50g IBr in 120 g of water. Assume 100% ionization. KBr = k1+Br1-.

3. Find the freezing point of solution composed of 74.15g MG(NO3)2 in 90 g of water. Assume 100% ionization.

4. Calculate the freezing point of a solution containing 12.0g of sucrose (C12H22O11) in 120g of water.

5. Calculate the boiling point of a solution containing 42.00 g of CuSO4 in 300g of water. CuSO4=Cu2+ + SO42-.

6. Find the molecular mass of a non-ionizing solute if 72g dissolved in 300g of water raise the boiling point to 100.860 degrees Celsius.

7. Find the molecular mass of a non-ionizing solute if 87.85g dissolved in 500g of water lower the freezing point to -3.00 degrees Celsius.

1. Find the boiling point of a solution composed of 110g of HgCl2 (a non-ionizing solute) in 175g of water.


mols HgCl2 = grams/molar mass
Substitute and solve for mols.

molality = m = mols/Kg solvent
Substitute and solve for m

delta T = i*Kb*m
Your post says i = 1
Kb = 1.51
Substitute and solve for delta T, then add to 100 C to find the new boiling point.

2. Find the freezing point of a solution composed of 50g IBr in 120 g of water. Assume 100% ionization. KBr = k1+Br1-.


I assume you made a typo and IBr really is KBr.
mols KBr = grams/molar mass.
Solve for mols.

molality = m = mols/kg solvent
Solve for m

delta T = i*Kf*m
i for KBr = 2
Kf = 1.86
Solve for delta T and subtract from 0 C to find the new freezing point.
delta T = i*Kf*m
i = 1

3,5,5 look just like 1 and 2.

6 and 7 are done by working backward.
delta T = i*Kf or Kb *m
You're given delta T, solve for m.

Then m = mols/kg solvent.
Solve for mols

Then mols = grams/molar mass. You're given mols and grams; solve for molar mass.
i = 1 for both 6 and 7

To solve these problems, you need to use the equations and formulas related to colligative properties, specifically boiling point elevation and freezing point depression.

1. Boiling point elevation:
The formula for boiling point elevation is ΔTb = Kb * m * i. Here, ΔTb represents the change in boiling point, Kb is the molal boiling point elevation constant of the solvent (water in this case), m is the molality of the solute (HgCl2), and i is the van't Hoff factor (1 for non-ionizing solutes).

2. Freezing point depression:
The formula for freezing point depression is ΔTf = Kf * m * i. Here, ΔTf represents the change in freezing point, Kf is the molal freezing point depression constant of the solvent (water), m is the molality of the solute (IBr), and i is the van't Hoff factor (2 for 100% ionization).

3. Freezing point depression:
The formula for freezing point depression is the same as mentioned in question 2.

4. Freezing point depression:
Use the same formula as in question 2 and 3.

5. Boiling point elevation:
Use the same formula as in question 1.

6. Boiling point elevation:
To determine the molecular mass of the solute, use the formula ΔTb = Kb * m * i. Rearrange the equation to solve for m, where m = (ΔTb) / (Kb * i). Then divide the mass of the solute (72g) by the calculated molality and multiply by the molar mass of water to find the molecular mass of the solute.

7. Freezing point depression:
To determine the molecular mass of the solute, use the formula ΔTf = Kf * m * i. Rearrange the equation to solve for m, where m = (ΔTf) / (Kf * i). Then divide the mass of the solute (87.85g) by the calculated molality and multiply by the molar mass of water to find the molecular mass of the solute.