At what temperature would a 1.75m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86 C/m.
-6.18 Celcius
delta T = Kf*i*m
You know Kf, i, and m. Calculate delta T and subtract from the normal freezing point of water which is zero degrees C.
To find the freezing point of the NaCl solution, we can use the equation for freezing point depression:
ΔTf = Kf * m * i
Where:
ΔTf = freezing point depression
Kf = molal freezing point depression constant
m = molality of the solution
i = van't Hoff factor
In this case, we are given the van't Hoff factor for NaCl as 1.9 and the molal freezing point depression constant (Kf) for water as 1.86 C/m.
First, we need to calculate the molality (m) of the NaCl solution. Molality is defined as the number of moles of solute (NaCl) per kilogram of solvent (water).
Given that the solution has a concentration of 1.75m NaCl, we know that 1.75 moles of NaCl are dissolved in 1 kilogram of water.
So, the molality (m) would be 1.75 mol of NaCl / 1 kg of water = 1.75 m.
Now, we can plug the given values into the freezing point depression equation:
ΔTf = (1.86 C/m) * (1.75 m) * (1.9)
ΔTf = 1.86 * 1.75 * 1.9
ΔTf = 5.5105 C
Therefore, the freezing point of the NaCl solution would be decreased by approximately 5.51 degrees Celsius.
To calculate the freezing point of a solution, we can use the equation:
ΔTf = Kf * m * i
Where:
ΔTf is the change in freezing point
Kf is the cryoscopic constant (freezing point depression constant) for the solvent
m is the molality of the solute
i is the van't Hoff factor
In this case, we want to find the freezing temperature, so we need to solve for ΔTf and then subtract it from the normal freezing temperature (0 degrees Celsius) to get the freezing point.
Given:
Kf for water (solvent) = 1.86 C/m
m (molality) = 1.75 m
i (van't Hoff factor for NaCl) = 1.9
First, calculate ΔTf using the equation:
ΔTf = Kf * m * i
ΔTf = 1.86 * 1.75 * 1.9
Now, substitute the given values and calculate:
ΔTf = 6.2115 C
Finally, subtract ΔTf from the normal freezing point (0 C) to get the freezing temperature:
Freezing temperature = 0 - ΔTf
Freezing temperature = 0 - 6.2115
Freezing temperature ≈ -6.21 C
Therefore, the 1.75m NaCl solution would freeze at approximately -6.21 degrees Celsius.