# Question #410dd

Feb 24, 2017

A) $2$

#### Explanation:

The van't Hoff factor, $i$, tells you the ratio that exists between the number of moles of solute dissolved in water and the number of moles of particles of solute produced in solution.

$i = \text{what you get in solution"/"what you dissolve to make the solution}$

Potassium hydroxide, $\text{KOH}$, is a soluble ionic compound, which means that it dissociates completely to form potassium cations and hydroxide anions

${\text{KOH"_ ((aq)) -> "K"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

Notice that every $1$ mole of potassium hydroxide that dissolves in solution produces

• one mole of potassium cations, $1 \times {\text{K}}^{+}$
• one mole of hydroxide anions, $1 \times {\text{OH}}^{-}$

This means that for every mole of potassium hydroxide you dissolve in solution, you get $2$ moles of particles of solute.

Therefore, the van't Hoff factor will be equal to $2$

$i = \left(1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mole K"^(+)))) + 1 color(red)(cancel(color(black)("mole OH"^(-)))))/(1color(red)(cancel(color(black)("mole KOH}}}}\right) = 2$

You can thus say that

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{i = 2}}} \to$ you get twice as many moles of particles of solute in solution than the number of moles of solute you dissolved to make the solution