Do volatile solutes affect the vapour pressure of solutions?

Question

If a non-volatile solute is added to water it will reduce vapour pressure as some space is occupied by non-volatile solute but what if the solute density is higher than water does vapour pressure still decrease?

Answer 1

The covered surface explanation is a kinetic explanation while the net evaporation is still ongoing. It does not provide the thermodynamic explanation when at equilibrium. While it slows down evaporation, it also slows down condensation.

Thermodynamic explanation of decreasing of vapor pressure of a compound with its molar fraction is related to its chemical potential.

$$\mu_i = \left(\dfrac{\partial G }{ \partial n_i}\right)_{p,T,n_j,j \ne i} \tag{1}$$

This potential decreases for ideal solutions with the molar fraction

$$\mu_{i,\text{(l)}} = \mu_{i,\text{(l)}}^{\circ} + RT \ln {x_i} \tag{2}$$

and so does the vapor pressure for ideal gas:

$$\mu_{i,\text{(g)}} = \mu_{i,\text{(g)}}^{\circ} + RT \ln {\left(\frac{p_i}{p_{i,0}}\right)} \tag{3}$$

For equilibrium, there must be valid

$$\mu_{i,\text{(l)}} = \mu_{i,\text{(g)}} \tag{4}$$ $$\mu^{\circ}_{i,\text{(l)}} = \mu^{\circ}_{i,\text{(g)}} \tag{4a}$$

and therefore

$$ RT \ln {x_i} = RT \ln {\left(\frac{p_i}{p_{i,0}}\right)} \tag{5}$$

From that is implied for ideal solutions

$$p = p_0 \cdot x_i \tag{6}$$