Showing posts with label van't Hoff factor. Show all posts
Showing posts with label van't Hoff factor. Show all posts

2013-07-02

Colligative properties and gas law problems 2013-07-02

1. 12.64g of sodium sulfate is dissolved in 400.0mL of water. What are the boiling point and freezing point of the solution?
Na2SO4 = 142.041g/mol
12.64g / 142.041g/mol = 0.0889884mols Na2SO4
{NOTE: don't round sig figs in the middle of a problem...}
(0.0889884mols Na2SO4) / 0.4000kg water = 0.22247m Na2SO4
{NOTE: density of water is 1.0000g/mL...}
ΔTbp = (0.512 ºC/m)(0.22247m)(3mol particles / 1mol Na2SO4) = 0.342ºC
{NOTE: I think I used a different value for the boiling point elevation constant in class. This one is correct.}
{NOTE: When Na2SO4 dissolves in water, the result is 2 Na+(aq) ions and 1 SO4-2(aq) ion. 3 particles.}
Tbp = 100.000ºC + 0.342ºC = 100.342ºC

ΔTfp = (1.858 ºC/m)(0.22247m)(3mol particles / 1mol Na2SO4) = 1.240ºC
Tfp = 0.000ºC – 1.240ºC = -1.240ºC

2. A weather balloon is filled with 23.65L of an ideal gas at 28.73ºC and 1.042atm pressure. It is released and rises to an altitude where the temperature is -6.35ºC the pressure is 0.842atm. How many moles of gas are in the balloon and what is its volume when it reaches the described altitude?
Plugging in to the Ideal Gas Law:
PV = nRT
(1.042atm)(23.65L) = n(0.08206 L.atm/mol.K)(301.88K)
n = 0.9948 moles
{NOTE: Convert to kelvins! If you pay attention to the units on R you'll be less likely to forget...}

Now that we know how many moles of gas are present, the next part could be solved by plugging in to the regular Ideal Gas Law again:
PV = nRT
(0.847atm)V = (0.9948mols)(0.08206 L.atm/mol.K)(266.80K)
V = 25.7L
Or by using the comparative form of the Ideal Gas Law:
P1V1/n1T1 = P2V2/n2T2
{NOTE: Since “n” is not changing, we can drop it from the equation...
(1.042atm)(23.65L) / 301.88K = (0.847atm)V2 / 266.80K
V2 = 25.7L


Practice, practice, practice...

2013-05-14

Email question - van't Hoff factor

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Can you please explain how to determine the i factor.  Now matter how much I read the book and look at examples I can't figure it out.  Thanks,
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There are a couple things that could be "the i factor", but I'll assume this question is about the van't Hoff factor that's used when determining changes due to colligative properties. Remember, colligative properties are properties of a solution that depend only upon the number of solute particle, not the identity of those solute particles. It doesn't matter if those solute particles are molecules, or cations, or anions, or a big random mixture of them all. When an ionic compound dissolves in water, it dissociates (at least somewhat) into its component ions. The number of ions (particles) that each formula unit breaks into when it dissolves is the van't Hoff factor. For example, if NaCl(s) is dissolved in water, it forms 1 Na+1(aq) ion and 1 Cl-1(aq) ion, so each formula unit (NaCl) forms 2 particles (ions) in solution; the van't Hoff factor is "2". For polyatomic ions, the ions don't break down into their individual atom, they stay polyatomic ions, so if K2SO4(s) is dissolved in water, it forms 2 K+1(aq) ions and 1 SO4-2(aq) ion in solution; the van't Hoff factor is "3".