# Question 7517e

May 14, 2017

$i = \frac{5}{4}$

#### Explanation:

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

$i = \text{total number of moles of particles of solute"/"number of moles of solute dissolved in solution}$

Now, you know that 50% of the electrolyte dissociates completely to produce ions

${\text{AB"_ ((aq)) -> "A"_ ((aq))^(+) + "B}}_{\left(a q\right)}^{-}$

Notice that ever mole of $\text{AB}$ that dissociates produces $1$ mole of ${\text{A}}^{+}$ and $1$ mole of ${\text{B}}^{-}$, or

$\text{1 mole A"^(+) + "1 mole B"^(-) = color(red)(2)color(white)(.)"moles ions}$

If we take $x$ to be the total number of moles of $\text{AB}$ dissolved in solution, you can say that

• $\textcolor{b l u e}{x \cdot 0.5} \to$ the number of molecules that dissociate
• $\textcolor{p u r p \le}{x \cdot 0.5} \to$ the number of molecules that dimerise

For every mole of $\text{AB}$ that dissociates, you have $\textcolor{red}{2}$ moles of ions, so the solution will contain

$\textcolor{red}{2} \cdot \textcolor{b l u e}{x \cdot 0.5} = x \to$ the number of moles of ions

Now, the other 50% of the molecules dimerise, i.e. the bond together to form a dimer.

"AB"_ ((aq)) + "AB"_ ((aq)) -> ("AB")_ (2(aq))#

This means that for every $1$ mole of molecules that dimerise, you get $\frac{1}{2}$ moles of dimers in the resulting solution, so

$\frac{1}{2} \cdot \textcolor{p u r p \le}{x \cdot 0.5} = \left(\frac{1}{4} \cdot x\right) \to$ the number of moles of dimers

This means that the total number of moles of particles of solute present in the solution will be

$x + \frac{1}{4} \cdot x = \frac{5}{4} \cdot x$

This means that the van't Hoff factor will be

$i = \frac{\frac{5}{4} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}} = \frac{5}{4}$