Equimolal solutions of #"KCl"# and compound #"X"# in #"H"_2"O"# show depressions of freezing pts. in a #4:1# ratio . Assuming #"KCl"# to be ionized completely , the compound #"X"# must ?
A) dissociate to extent of 50 %
B) Hydrolyse to the extent of 80 %
C) Trimerize to extent of 75 %
D) Dimerize to extent of 50%
A) dissociate to extent of 50 %
B) Hydrolyse to the extent of 80 %
C) Trimerize to extent of 75 %
D) Dimerize to extent of 50%
1 Answer
Here's my take on this.
Explanation:
The idea here is that since freezing-point depression is a colligative property, it will depend exclusively on how many particles of solute are present in solution, and not on the nature of those particles.
The equation that allows you to calculate freezing-point depression looks like this
#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "# , where
In your case, the solutions are equimolal, which means that they have equal molalities.
Since the solvent is the same in both cases, you can conclude that the difference between the freezing-point depression of the solution that contains potassium chloride,
You will have
#DeltaT_"f KCl" = i_"KCl" * K_f * b#
#DeltaT_"f X" = i_"X" * K_f * b#
Divide these two equations to get
#(DeltaT_"f KCl")/(DeltaT_"f X") = (i_"KCl" * color(red)(cancel(color(black)(K_f * b))))/(i_"f X" * color(red)(cancel(color(black)(K_f * b))))#
But since you know that the freezing-point depressions are in a
#(DeltaT_"f KCl")/(DeltaT_"f X") = i_"f KCl"/i_"f X" = 4/1#
This is equivalent to
#color(green)(|bar(ul(color(white)(a/a)color(black)(i_"f X" = 1/4 xx i_"f KCl")color(white)(a/a)|)))#
Now, the van't Hoff factors 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 that are created in solution when the solute dissolves.
Potassium chloride dissociates to form potassium cations,
#"KCl"_ ((aq)) -> "K"_ ((aq))^(+) + "Cl"_((aq))^(-)#
Since every mole of potassium chloride that dissolves in solution produces one mole of potassium cations and one mole of chloride anions, the van't Hoff factor for this compound will be
#i_"f KCl" = 2#
This means that the van't Hoff factor of compound
#i_"f X" = 1/4 xx 2 = 0.5#
So, when you dissolve one mole of compound
Right from the start, this should tell you that compound
Now, let's assume that you dissolve
A dimer is formed when two monomers, which in your case would be two molecules of
If
#n/2 -> # remain as molecules of#"X"#
#1/2 * n/2 = n/4 -># exist as dimers
In this case,
#n/2 + n/4 = (3n)/4 -># moles of particles of solute
Not what we're looking for. On the other hand, if
#n/4 -># remain as molecules of#"X"#
#1/3 * (3n)/4 = n/4 -># exist as trimers
In this case,
#n/4 + n/4 = n/2 -># moles of particles of solute
Therefore, the answer is (3) Trimerise to extent of