You need to find the freezing point of a 1.50m aqueous NaCl solution. You calculate []Tf to be 1.86 degrees C/m x 3.00m or 5.86 degrees C. What is the temperature at which the solution freezes?
The normal freezing point is zero C.
Therefore, the new freezing point will be -5.86 degrees C. However, when I multiply 3 x 1.86 I obtain 5.58 and the freezing point would be -5.58.
To find the freezing point of a solution, you can use the equation:
△Tf = kf x m
where △Tf is the freezing point depression, kf is the freezing point depression constant (a characteristic property of the solvent), and m is the molality of the solution. In this case, the solution has a molality (m) of 1.50m and the freezing point depression constant (kf) for water is 1.86 degrees C/m.
You have already calculated △Tf to be 5.86 degrees C (by multiplying kf with m). To find the temperature at which the solution freezes, you need to subtract △Tf from the freezing point of pure water, which is 0 degrees Celsius.
So, to determine the freezing point of the solution, you can use the equation:
Freezing point of solution = Freezing point of pure solvent - △Tf
Freezing point of solution = 0°C - 5.86°C
Simplifying, you get:
Freezing point of solution = -5.86°C
Therefore, the temperature at which the solution freezes is approximately -5.86 degrees Celsius.