What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 in 115 g of water? The molal freezing point depression constant for water is 1.86°C/m. (Note that when Ca(NO3)2 dissolves in water Ca2+ and NO32- ions are produced).

mols Ca(NO3)2 = grams/molar mass

molality = mols/kg H2O
delta T = i*Kf*m
You have Kf and m. i for Ca(NO3)2 is 3.
For freezing point, 0-delta T = f.p.

should i get -3.34? can't help but feel as though i did wrong

-3.34 C agrees with my quick calculation.

To find the freezing point of the solution, we need to use the formula for freezing point depression:

ΔT = Kf * m

Where:
ΔT is the change in freezing point
Kf is the molal freezing point depression constant
m is the molality of the solute

First, let's calculate the molality (m) of the solute (Ca(NO3)2).

Molality (m) is defined as the moles of solute per kilogram of solvent.

1. Calculate the moles of Ca(NO3)2:
We'll use the formula: moles = mass / molar mass.
The molar mass of Ca(NO3)2 is (40.08 g/mol * 1) + (14.01 g/mol * 2) + (16.00 g/mol * 6) = 164.09 g/mol.
So, moles of Ca(NO3)2 = 11.3 g / 164.09 g/mol = 0.0689 mol.

2. Calculate the mass of water in kg:
Mass of water = 115 g / 1000 = 0.115 kg.

3. Calculate molality (m):
Molality (m) = moles of solute / mass of solvent in kg.
Molality (m) = 0.0689 mol / 0.115 kg = 0.599 mol/kg.

Now, we can calculate the freezing point depression (ΔT) using the given molal freezing point depression constant (Kf).

ΔT = Kf * m
ΔT = 1.86°C/m * 0.599 mol/kg = 1.11514°C.

Finally, to find the freezing point of the solution, we subtract the calculated freezing point depression from the freezing point of pure water (0°C).

Freezing point = 0°C - 1.11514°C = -1.11514°C.

Therefore, the freezing point of the solution prepared by dissolving 11.3 g of Ca(NO3)2 in 115 g of water is approximately -1.11514°C.