Question #7517e

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 = "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

#"AB"_ ((aq)) -> "A"_ ((aq))^(+) + "B"_ ((aq))^(-)#

Notice that ever mole of #"AB"# that dissociates produces #1# mole of #"A"^(+)# and #1# mole of #"B"^(-)#, or

#"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 #"AB"# dissolved in solution, you can say that

• #color(blue)(x * 0.5) -># the number of molecules that dissociate
• #color(purple)(x * 0.5) -># the number of molecules that dimerise

For every mole of #"AB"# that dissociates, you have #color(red)(2)# moles of ions, so the solution will contain

#color(red)(2) * color(blue)(x * 0.5) = x -># 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 #1/2# moles of dimers in the resulting solution, so

#1/2 * color(purple)(x * 0.5) = (1/4 * x) -># 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 + 1/4 * x = 5/4 * x#

This means that the van't Hoff factor will be

#i = (5/4 * color(red)(cancel(color(black)(x))))/(color(red)(cancel(color(black)(x)))) = 5/4#