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?

Answer 1

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.