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...}

ΔT_bp = (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.}

T_bp = 100.000ºC + 0.342ºC = 100.342ºC

ΔT_fp = (1.858 ºC/m)(0.22247m)(3mol particles / 1mol Na₂SO₄) = 1.240ºC

T_fp = 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...