Vapor Pressure in fresh vs salt water

[physicsss]

For water at 30°C, when 100 g of pure water the vapor pressure is 31.82 mm Hg
and when 50g of NaCl dissolved in 100g of water vapor pressure is 27.68 mm Hg

The vapor pressure of pure water is different than the vapor pressure of water in a salt water solution. Explain why.

Can anyone get me started? I have no idea how to approach this...thanks.

 

[ShawnD]

For water at 30°C, when 100 g of pure water the vapor pressure is 31.82 mm Hg The vapor pressure of pure water is different than the vapor pressure of water in a salt water solution. Explain why.

The vapor pressure decreases due to cohesive forces between the salt and the water. If you pull out your periodic table of electronegativities, you will see that sodium chloride is considerably more polar than water is. Due to that extra polarity, water can make much stronger bonds to sodium chloride than it can to itself. Stronger inter/intramolecular forces in solution always cause the vapor pressure to decrease.

You can also have the opposite happen. If you add something nonpolar such as carbon disulphide, the vapor pressure will increase considerably since water and carbon disulphide don't attract each other that much.

 

[GCT]

You can also explain this in terms of entropy, try it yourself.

Compare the entropy of liquid vs gaseous state

Compare the entropy state of the solution v.s pure water

Which one would experience the greatest unfavorable entropy change due to vaporization?

 

[Gokul43201]

The vapor pressure decreases due to cohesive forces between the salt and the water. If you pull out your periodic table of electronegativities, you will see that sodium chloride is considerably more polar than water is. Due to that extra polarity, water can make much stronger bonds to sodium chloride than it can to itself. Stronger inter/intramolecular forces in solution always cause the vapor pressure to decrease.

You can also have the opposite happen. If you add something nonpolar such as carbon disulphide, the vapor pressure will increase considerably since water and carbon disulphide don't attract each other that much.

More important is the fact that the solute molecules make up the surface of the solution in hte same mole fraction as in the bulk. Since evaporation is a surface phenomenon, by reducing the fraction of volatile solvent molecules in the surface, you alter the liquid-vapor equilibrium. The more solute molecules you have in the surface, the fewer solvent molecules there are, and hence the lower is the evaporation rate. The equilibrium constant for evaporation, which is a linear function of the vapor pressure is merely the ratio of the forward and reverse rate constants. By lowering the forward rate constant (while fixing that for the reverse), you reduce the vapor pressure proportionally. This explains whay Raoult's Law is approximately correct.

The ionic character of the solute has only a weak effect on the vapor pressure. That's why Raoult's Law works very well in low dilutions. The only factor that affects the vapor pressure is the mole fraction of the solvent, and no matter whether the solute is NaCl or CuSO4, a 0.01 mole fraction solution will see a 1% lowering of the vapor pressure.

Even the above argument uses some extent of hand-waviness, and the explanation based on entropy is far more elegant, in my opinion (but perhaps less intuitive, at the intructional level).