Does vapor pressure vary with the number of particles in solution? What if you have #"1 M"# of #"NaCl"#, vs. #"1 M"# #"KCl"#, vs. #"1 M"# #"CaCl"_2#?

Answer 1

Maybe the following concept map will make things clearer.

These electrolytes generate approximately two, two, and three ions, respectively, in solution for each formula unit of the solute.

#"NaCl"(aq) -> "Na"^(+)(aq) + "Cl"^(-)(aq)#

#"KCl"(aq) -> "K"^(+)(aq) + "Cl"^(-)(aq)#

#"CaCl"_2(aq) -> "CaCl"^(+)(aq) + "Cl"^(-)(aq)#
#"CaCl"^(+)(aq) -> "Ca"^(2+)(aq) + "Cl"^(-)(aq)#

Hence, the molar and molal concentration, and thus the NONIDEAL "mol fraction" (will be discussed below), of electrolyte particles originating from #bb("CaCl"_2)# will be higher from analogous solutions of #"NaCl"# and #"KCl"#.

The approximate concentrations of the above solutions are #"2 M"#, #"2 M"#, and #"3 M"# simply by counting the electrolyte particles and assuming #100%# dissociation.

Refer to this answer for more on osmotic pressure, vapor pressure, and freezing point. All of these depend on the effective concentration of electrolytes in the solution.


RAOULT'S LAW IS ONLY FOR IDEAL SOLUTIONS

The decrease in vapor pressure, #DeltaP_j = P_j - P_j^"*"#, can be determined based on Raoult's law,

#P_j = chi_(j(l))P_j^"*"#

in terms of mol fractions of solute #j# in the liquid phase, but the problem is that it is only for ideal solutions. Equations such as

#DeltaT_f = T_f - T_f^"*" = -iK_fm#

already account for nonideality using the van't Hoff factor #i#. Sometimes we assume what #i# is by looking at the number of particles are in solution per formula unit of solute, but in general #i# accounts for nonideality. For example, #i = 1.9# for #"NaCl"#, but #2.7# for #"MgCl"_2#.

AN ADJUSTED "RAOULT'S LAW"

Raoult's law can be rewritten in another way that does account for nonideality:

#P_j = a_(j(l))P_j^"*"#

where:

  • #a_(j(l)) = gamma_jchi_j# is the activity of the solute #j# in the liquid phase.
  • #gamma_j# is the activity coefficient of the solute #j#.
  • #chi_(j(l))# is the mol fraction of the solute #j# in the liquid phase.

The activity coefficient #gamma_j# can be considered as a weighting factor to account for nonideality, and the activity #a_(j(l))# can be considered the "nonideal" version of the mol fraction.

Since mol fraction, molality, and molarity are approximately interconvertible (see the above concept map), we can indirectly say that the "true mol fraction" (the activity) of solute in solution is not simply due to the original solute, such as #"CaCl"_2#, but due to ALL the electrolytes that it generates, namely #"Ca"^(+)#, #"Cl"^(-)#, AND #"CaCl"^(+)#!


So, no, we cannot say that the vapor pressure decrease is to the same extent for solutions of nonvolatile solutes, even if they have the same face-value concentrations.

We must look at their effective concentrations instead, which is not readily-seen trend and might have to be inferred indirectly when it comes to vapor pressure.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

Yes, vapor pressure does vary with the number of particles in solution. In a solution of 1 M NaCl, the vapor pressure will be lower compared to a solution of 1 M KCl, which in turn will be lower than a solution of 1 M CaCl2. This is because NaCl dissociates into two ions (Na+ and Cl-) in solution, KCl dissociates into two ions (K+ and Cl-), and CaCl2 dissociates into three ions (Ca2+ and 2Cl-), resulting in a higher concentration of solute particles in the latter case, leading to a greater reduction in vapor pressure according to Raoult's law.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7