Calculate the boiling points of freezing points of the following solutions in water.

a) boiling point of a 2.25m solution of propylene glycol (a non-electrolyte)
b)freezing point of a 1.0m solution of potassium iodide
c)freezing point of a 1.0m solution of nickel(II) nitrate
d)boiling point of a .50m solution of sodium chloride

Too many of the same kind of problem.

General procedure:
delta T = Kb*m for boiling point
delta T = Kf(m for freezing point.
for b, c, and d, add an i to the formulas (i.e., delta T = i*Kf*m or delta T = i*Kb*m) where i stands for the number of particles. For b and d that is 2. For c it is 3.

To calculate the boiling points or freezing points of solutions, you can use colligative properties. Colligative properties are properties of a solution that depend on the concentration of solute particles, rather than their identity.

For boiling point elevation, the equation used is:

△T = K_b * m

Where:
△T is the change in boiling point,
K_b is the molal boiling point elevation constant of the solvent, and
m is the molality of the solution (moles of solute per kilogram of solvent).

For freezing point depression, the equation used is:

△T = K_f * m

Where:
△T is the change in freezing point,
K_f is the molal freezing point depression constant of the solvent, and
m is the molality of the solution.

Now let's calculate the boiling points and freezing points of the given solutions in water:

a) Boiling point elevation of a 2.25m solution of propylene glycol:
To find the boiling point elevation, you need the molal boiling point elevation constant (K_b) for water. For water, K_b is approximately 0.512 degrees Celsius/m.

△T = K_b * m
△T = 0.512 * 2.25
△T = 1.152 degrees Celsius

So, the boiling point of the 2.25m propylene glycol solution would be 1.152 degrees Celsius higher than the boiling point of pure water.

b) Freezing point depression of a 1.0m solution of potassium iodide:
To find the freezing point depression, you need the molal freezing point depression constant (K_f) for water. For water, K_f is approximately 1.86 degrees Celsius/m.

△T = K_f * m
△T = 1.86 * 1.0
△T = 1.86 degrees Celsius

So, the freezing point of the 1.0m potassium iodide solution would be 1.86 degrees Celsius lower than the freezing point of pure water.

c) Freezing point depression of a 1.0m solution of nickel(II) nitrate:
Again, using the freezing point depression equation, with the appropriate K_f value for water.

△T = K_f * m
△T = 1.86 * 1.0
△T = 1.86 degrees Celsius

Therefore, the freezing point of the 1.0m nickel(II) nitrate solution would be 1.86 degrees Celsius lower than the freezing point of pure water.

d) Boiling point elevation of a 0.50m solution of sodium chloride:
Using the boiling point elevation equation, with the appropriate K_b value for water.

△T = K_b * m
△T = 0.512 * 0.50
△T = 0.256 degrees Celsius

Hence, the boiling point of the 0.50m sodium chloride solution would be 0.256 degrees Celsius higher than the boiling point of pure water.