Question #ecc6a

In simple terms, you can think about the van't Hoff factor, `i`, as a sort of bang for your buck thing because it tells you how many moles of particles of solute are present in the solution for every mole of solute dissolved to make the solution.

`i = "moles of particles of solute in solution"/"moles of solute dissolved to make the solution"`

Now, sodium phosphate, `"Na"_3"PO"_4`, is a soluble ionic compound, which means that it dissociates completely in aqueous solution to produce sodium cations, `"Na"^(+)`, and phosphate anions, `"PO"_4^(3-)`.

`"Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-)`

Notice that for every mole of sodium phosphate that you dissolve in water, you get a total of

`"3 moles Na"^(+) + "1 mole PO"_4^(3-) = "4 moles ions"`

This means that the predicted value of the van't Hoff factor will be

`i = (4 color(red)(cancel(color(black)("moles ions"))))/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) color(white)(aacolor(black)(larr " "color(blue)("what you get in solution")color(white)(aaaa))/(color(black)(larr" "color(blue)("what you dissolve in solution"))`

`i = 4`