Question #a0387

The idea here is that you're going to observe a smaller molar mass for silver nitrate because some of the salt will dissociate when dissolved in water to produce silver(I) cations and nitrate anions in aqueous solution.

#"AgNO"_ (3(aq)) -> "Ag"_ ((aq))^(+) + "NO"_ (3(aq))^(-)#

Notice that every mole of silver nitrate that dissociates in solution produces #1# mole of silver(I) cations and #1# mole of nitrate anions, hence #color(red)(2)# moles of ions. Keep this in mind.

Now, let's say that your sample has a mass of #m# #"g"# and contains #n# moles of silver nitrate.

You can express the molar mass of the salt, which represents the mass of exactly #1# mole of silver nitrate, in terms of the mass #m# and the number of moles #n#

#M_M = m/n#

You calculate that silver nitrate has a molar mass of #"170 g mol"^(-1)#, so you can say that

#170 color(red)(cancel(color(black)("g mol"^(-1)))) = (m color(red)(cancel(color(black)("g"))))/(n color(red)(cancel(color(black)("moles"))))#

#170 = m/n" "color(orange)("(*)")#

At this point, you dissolve the sample in water. Let's say that #f# represents the fraction of moles of silver nitrate that dissociate to produce ions.

You can say that the solution will contain

#1 - f * n -># the number of moles of silver nitrate that do not dissociate

#color(red)(2) * f * n -># the number of moles of ions produced in solution

The total number of moles of solute, dissociate and undissociated, present in the solution is equal to

#1 - fn + color(red)(2) * fn = 1 + fn#

At this point, you observe that the molar mass comes out to be

#92.64 color(red)(cancel(color(black)("g mol"^(-1)))) = (m color(red)(cancel(color(black)("g"))))/( (1 + fn)color(red)(cancel(color(black)("moles"))))#

#92.64 = m/(1 + fn)" "color(darkorange)("(* *)")#

Notice that we have two equations with three unknowns, the mass of the sample, the number of moles it contains, and the fraction that dissociates.

In order to be able to calculate the value of #f#, we will pick a sample of silver nitrate to work with, then use equation #color(orange)("(*)")# to find the number of moles it contains.

To make the calculations easier, let's say that we're working with #"170 g"# of silver nitrate. Since this represents the molar mass of the salt, you can say that

#"170 g AgNO"_3 " " -> " " "1 mole AgNO"_3#

At this point, you will have

#{(m= "170 g"), (n = "1 mole") :}#

Plug this into equation #color(darkorange)("(* *)")# to find the value of #f#

#92.64 = 170/(1 + f * 1)#

#1 + f = 170/92.64#

#f = 170/92.64 - 1 = 0.835#

So, you know that a fraction equal to #0.835# of the salt dissociates. To convert this to a percentage, multiply the value of #f# by #100%#.

#color(darkgreen)(ul(color(black)("% dissociation" = 83.5%)))#