# How would you predict the boiling point of a 0.200 m solution of potassium iodide in water?

Jun 15, 2017

${100.205}^{\text{o""C}}$

#### Explanation:

To predict the change in boiling point of a solution from the original boiling point, we can use the equation

$\Delta {T}_{\text{b" = i · m · K_"b}}$

where

• $\Delta {T}_{\text{b}}$ is the (positive quantity) change in boiling point of the solution, in $\text{^"o""C}$

• $i$ is called the van't Hoff factor, which for these purposes is just the number of dissolved ions per unit of solute, which is $2$ (${\text{K}}^{+}$ and ${\text{I}}^{-}$)

• $m$ is the molality of the solution, which in this case is $0.200$ $m$

• ${K}_{\text{b}}$ is the molal boiling point constant of the solvent (water), which, referring to a table for these constants, is $0.512 \frac{\text{^"o""C}}{m}$.

Plugging in these known quantities to the equation, we have

$\Delta {T}_{\text{b" = (2)(0.200cancel(m))(0.512(""^"o""C")/(cancel(m))) = 0.205^"o""C}}$

This represent the change in temperature of the boiling point of the solution, NOT the actual boiling point. To find the new boiling point, we simply add this to the regular boiling point of water, ${100}^{\text{o""C}}$:

${100}^{\text{o""C" + 0.205^"o""C" = color(red)(100.205^"o""C}}$

The new boiling point of the solution is thus color(red)(100.205^"o""C"