Question

The boiling point of an aqueous solution is 102.42 °C. What is the freezing point? Constants can be found here.

Answer 1

Since pure water boils at 100.00°C, the change in boiling point is
ΔTb = (102.42 – 100.00)°C = 2.42 °C.
The molal concentration, m (or b), and the van\'t Hoff factor, i, are constant, so solve the boiling-point-elevation formula for m·i.
\[\Delta T _{b} = k _{b} \times m \times i\]
\[mi = \frac{\Delta T _{b} }{ k _{b} }\]
Then substitute that expression for m·i in the freezing-point-depression formula and solve for \[T _{f}\]