What is the freezing point of an aqueous solution that boils at 103.7 degrees celsius?

For this, we use equations from the colligative properties of solutions specifically boiling point elevation and freezing point depression. The equations for these are expressed as:

ΔTb = kb m

where k is a boiling point elevation constant and m is the concentration in terms of molality

ΔTf = kf m 

where k is a freezing point elevation constant and m is the concentration in terms of molality

We use both expression to solve for the freezing point. For this case, concentration is the same. The equation will then be:

ΔTf = kf ( ΔTb / kb )
0-Tf = 1.86 (103.7 - 100 / 0.512 )
Tf = -13.4°C

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