Question

1. Calculate the vapor pressure of a solution made by dissolving 65.0 g CaCl2, C6H12O6, in 655 g of water. The vapor pressure of pure water is 47.1 torr at 37°C.

2. What is the vapor pressure of an aqueous solution that has a solute mole fraction of 0.2000? The vapor pressure of water is 35.576 mmHg at 25 °C.

Answer 1

1.

To use Raoult's Law , we need to calculate the mole fraction of water (the solvent) in this salt-water solution.

P_solution = X_solvent * P⁰_solvent

$X_{solvent} = \frac {"molesof water"}{molesofsolute+molesofsolvent}$

$X_{solvent}= \frac {n_{}water}{n_{solute}+n_{water}}$

The molar mass of CaCl₂ is 111 g/mol and of water is 18 g/mol. So,

$n_{water}= \frac {655g}{18 g/mol}= 36,39mol$

and

$n_{solutes}= \frac{65g}{111g/mol}=0,586mol$

but this is really:

n_Ca²+ = 0,586 mol

n_Cl- = 1,172 mol

n_water = 36,39mol

$X_{solvent}= \frac {36,39}{0,586+1,172+36,39}=0,954$

Note that this still relatively dilute. The pressure of the solution is then calculated via Raoult's Law

P_solution = 0.954 × 47.1= 44.93torr

2.

χ_solv = 1.0000 − 0.2000 = 0.8000 <--- solv means solvent

Use Raoult's Law:

P_solution = (χ_solvent) (P⁰_solvent)

x = (0.8000) * (35.576)

x = 28.46 mmHg