Why is water evaporated from the ocean not salty?

Water molecules are polar; this basically means that they have a "positive side" and a "negative side"

Salt is composed by Na$^+$ and Cl$^-$ ions held together by electrostatic forces (is a ionic compound). When salt is put into water, it dissociates, i.e. the Na$^+$ ions are separated from the Cl$^-$ ions.

When in the water, such ions are surrounded by water molecules facing them with the side which has charge opposite to that of the ion; this is because this way they can reach a lower energy state, being their electrostatic field screened by that of the water molecules (picture below [source]).

Water evaporates when the thermal energy of the molecules is high enough to break about half the hydrogen bonds between them [source]. For the ions, it is much more difficult to evaporate, because their thermal energy would have to be enough to compensate the effect of the water molecules which surround them.

Basically, both the water molecules and the ions are in what is called a potential energy well: to "kick them out" of the well, we have to provide them with an energy as high as the depth of the energy well $\Delta E$ (picture below).

The depth of the well in which the water molecules are (due to hydrogen bonding) is much lower than the depth of the well where the ions are. So, a far higher thermal energy is needed to take an ion out of the water. Since thermal energy is proportional to $k_B T$, where $k_B$ is Boltzmann's constant, this means that a far higher temperature is needed.

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Update: Some numbers

To get an idea of the order of magnitudes of the energies involved, we should consider the following:

• At room temperature ($T_r\simeq298$K), $k_B T_r = 0.026$ eV (however, we should keep in mind that this is just an order of magnitude...)
• The energy of an hydrogen bond (hydrogen bond enthalpy) in water is around $23.3$ kJ/mol = $0.24$ eV = $9.3 \ k_B T_r$, and in order a volume for water to evaporate, about half of all the hydrogen bonds in the volume must be broken:

There is no standard  definition for the hydrogen bond energy. In liquid water, the energy of attraction between water  molecules (hydrogen bond enthalpy) is optimally about $23.3$ kJ/mol (Suresh and Naik, 2000)  and  almost five times the average thermal collision fluctuation at $25$°C. This is the energy required for  breaking and completely separating the bond, and equals about half the enthalpy of vaporization ($44$ kJ/mol  at $25$°C), as an average of just under two hydrogen bonds per molecule are broken when water  evaporates. [source]

• The energy gained by putting an ion in water (technical term "hydrating" the ion) is known as the hydration energy or hydration enthalpy ($\Delta H_{hyd}$). Since we are interested in the opposite process (the remotion or de-hydration of the ion), we have to take $-\Delta H_{hyd}$. Here we can find some numbers. We can see that

$$\Delta H_{hyd}(\text{Na}^+) = -406 \ \text{kJ/mol} = -4.2 \ \text{eV} = -162\ k_B T_r$$

$$\Delta H_{hyd}(\text{Cl}^-) = -363 \ \text{kJ/mol} = -3.8 \ \text{eV} = -145 \ k_B T_r$$

At any given temperature and pressure, liquids (like water) and solids (like salt) are in equilibrium with their respective gaseous form. The vapour pressure describes this phenomenon. Often, solids have a lower vapour pressure as liquids. The higher the vapour pressure, the more volatile a compound is. And in the instance of pure sodium chloride / NaCl, this vapour pressure is much lower than the one of water. This alone contributes why heating sea water, and condensing the vapour, will first yield water.

In addition, to perform such a distillation, the Raoult's law applies, too. Basically, there is a total pressure $p$ in the system observed that is the sum of the individual partial pressures of the two components (water and salt). The partial pressure takes into account the vapour pressure the pure component possess (at this temperature and pressure) times the concentration of this component in the mixture (expressed as molar fraction):

$p (\mbox{total}) = p(\mbox{water}) * x(\mbox{water}) + p(\mbox{NaCl}) * x(\mbox{NaCl})$

where the molar fractions will add up to one. So if we simplify "seawater" as a solution of NaCl in water:

$x(\mbox{water}) + x({\mbox{NaCl}}) = 1$

The contribution of the total pressure $p$ by NaCl is lowered further as at ambient conditions, water may dissolve only 359 g NaCl per litre of water (reference), hence excluding a solution of higher concentration of NaCl than about 36 mass%.

This is a simplified view, leaving out any idea of "molecules" or "ions". Raoult's law is an idealization, too (hence, for example, azeotropes like ethanol/water). Of course, sea water contains more than NaCl; and to remove the later from the former, reverse osmosis is an energetically more favourable technique. Eventually, to distill NaCl, you need conditions of very high temperature and very low pressure, far distant from the ones to distill water.

Alcohol will boil off first
The boiling point is 78.37 °C
There will be some water but the ratio of alcohol in the vapor phase will be higher than the liquid. Water and alcohol are azeotropes so things flop at about 70% alcohol.
The size is not that much of a factor - hexane is volatile. wiki

Table salt (NA CL) is not much more volatile than a rock

melting point 801 °C (1,474 °F)
boiling point 1,413 °C (2,575 °F)
wiki

The lowest vapor pressure I could find is at 759.88 °K = 486.73 °C
1.88322997019E-006 kPa
Vapor Pressure of Sodium chloride

The vapor pressure of water is 101.32 kPa at 100 °C

Water is 53,801,182 times more volatile at 100 °C than salt at 486.73 °C

Salt is dissolved in water but that does not make the salt more volatile.

Basically no salt evaporates. With very rapid boiling you might get some entrainment. The vapor phase is going to have essentially not salts or minerals as they are solids are at 100 °C.

Take a quart of water and add a cup of salt. Put it on the stove and let all the water boil off. You will have a cup of salt.