At what temperature would 1.3 m NaCl freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86 deg C/m.
delta T = i*Kf*m
delta T = 1.9*1.86*1.3 = ?
Subtract from zero to find new freezing point.
To determine the freezing point of a solution, you can use the equation:
ΔT = Kf * m * i
Where:
- ΔT is the change in freezing point
- Kf is the cryoscopic constant (freezing point depression constant) for the solvent (water in this case)
- m is the molality of the solute (NaCl in this case)
- i is the van't Hoff factor (the number of particles formed in the solution per molecule of solute)
First, we need to calculate the molality (m):
Molality (m) = moles of solute / kg of solvent
Given that you have 1.3 moles of NaCl and assuming the molecular weight of NaCl is 58.44 g/mol, we can calculate the mass (in kg) of solvent (water):
mass of solvent (water) = moles of solvent (water) * molar mass of water
Assuming the molar mass of water is 18.015 g/mol, the moles of water in 1.3 m NaCl would be:
moles of water = 1.3 mol * (1 / 1.9) = 0.6842 mol
mass of water = 0.6842 mol * 18.015 g/mol = 12.334 g ≈ 0.01234 kg
Next, we can calculate the molality (m):
m = moles of solute / kg of solvent
m = 1.3 mol / 0.01234 kg ≈ 105.32 mol/kg
Now that we have the molality (m) and the van't Hoff factor (i = 1.9), we can calculate the change in freezing point (ΔT):
ΔT = Kf * m * i
ΔT = 1.86 °C/m * 105.32 mol/kg * 1.9
Calculating this gives us:
ΔT ≈ 361.5 °C
To determine the freezing point of the solution, subtract the change in freezing point from the normal freezing point of water (0 °C):
Freezing point of the solution = 0 °C - 361.5 °C ≈ -361.5 °C
Therefore, the freezing point of the 1.3 m NaCl solution would be approximately -361.5 °C.