Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

How can I determine the van't Hoff factor of a substance from its formula?

Here's how you do it.

The van't Hoff factor, i, is the number of particles formed in a solution from one formula unit of .

Notice that i is a property of the solute. In an ideal solution, i does not depend on the concentration of the solution.

For a nonelectrolyte

If the solute is a nonelectrolyte (i.e. it does not separate into ions in solution), i = 1

For example, "sucrose(s) → sucrose (aq)".

i = 1, because 1 molecule of sucrose forms only one particle in solution.

For a strong electrolyte

If the solute is a strong electrolyte (i.e. it separates into ions in solution), i > 1.

Some examples are:

"NaCl(s)" → "Na"^+("aq") + "Cl"^"-"("aq"); i = 2

One formula unit of "NaCl" will form two particles in solution, an "Na"^+ ion and a "Cl"^"-" ion.

"CaCl"_2(s) → "Ca"^"2+"("aq") + "2Cl"^"-"("aq"); i = 3

One formula unit of "CaCl"_2 will form three particles in solution, a "Ca"^"2+" ion and two "Cl"^"-" ions.

Here's another example:

"Fe"_2("SO"_4)_3("s") → "2Fe"^"3+"("aq") + "3SO"_4^"2-"("aq"); i = 5

For a weak electrolyte

If the solute is a weak electrolyte , it dissociates only to a limited extent.

For example, acetic acid is a weak acid. We often set up an ICE table to calculate the number of particles in a 1 mol/L solution.

color(white)(mmmmmm)"HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^"-" "I/mol·L"^"-1":color(white)(ml) 1color(white)(mmmmmmml) 0color(white)(mmm) 0 "C/mol·L"^"-1": color(white)(m)"-"xcolor(white)(mmmmmm) +xcolor(white)(m)+x "E/mol·L"^"-1":color(white)(l) 1-xcolor(white)(mmmmmm)xcolor(white)(mmm) x

At equilibrium, we have 1-xcolor(white)(l) "mol of HA", xcolor(white)(l) "mol of H"_3"O"^+, and xcolor(white)(l) "mol of A"^"-".

"Total moles" = (1-x + x + x)color(white)(l) "mol" = (1+x)color(white)(l) "mol", so i = 1+x".

Usually, x < 0.05, so i < 1.05 ≈ 1.