I've seen some formulas around in other questions and Google searches, but my chemistry is pretty much dead so I have no clue where to find the relevant values to calculate it myself. I just need to make an oil bath at 104 °C, but I would prefer to use a water solution instead since it is cheaper and easier to clean up. Doesn't need to be salt, can be anything commonly available at your average household.


According to [1, pp. 281–282], solution of sodium chloride $\ce{NaCl}$ prepared by dissolving 25 g of salt in 100 g of water has boiling point of 104.6 °C.

Additional data is available in the following table for the aqueous solutions of common salts and bases. English transcription; column 1: Compound; columns 2–6: Concentration, g/100 g water — boiling points $(t_\mathrm{b.p.},~\pu{°C})$.

Boiling points of water solutions — p. 281

Boiling points of water solutions — p. 282

Note, however, that there is a significant drawback of using boiling salt solution due to the shift of boiling point upwards as the water boils off. You either have to add water to the mark from time to time, or use a bath with another heat carrier and a thermal sensor.


  1. V.A. Rabinovich and Z.Y. Havin. (Eds.) "Kratkii khimicheskii spravochnik" (Brief chemical handbook), 2nd ed., Khimiya, Leningrad, 1978. (in Russian)
  • 6
    $\begingroup$ Consider the altitude at which you will run your operation, too. @Louis Victor. $\endgroup$
    – Alchimista
    Jun 3 '19 at 10:03
  • 1
    $\begingroup$ Wow, lithium chloride is powerful stuff! $\endgroup$
    – Curt F.
    Jun 3 '19 at 20:42

Boiling point elevation of water by adding salt can be predicted with the formula:

$$\Delta T_b= K_b\, b_{solute} \, i$$

We look for $\Delta T_b=4$, using water $K_b=0.512$, and $i=2$ for salt $\ce{NaCl}$

$b$ is the molality, and an estimate $b=\pu{4 mol/kg}$ as the number $\ce{NaCl}$ moles per kilogram of solution. Then

$$ \begin{align} \frac{g_{salt}}{100 g_{H2O}} &= \frac{b\times100}{(1000-b\times PM_{salt})}\, PM_{salt} \\ \frac{g_{NaCl}}{100 g_{H2O}} &=\frac{4\times100}{(1000-4\times 58.5)} \times 58.5\\ &=30.55 \end{align} $$ (results).

$i$ of Different salts can be found (some examples here).


A pressure cooker will allow boiling temperature to increase to 115 degrees or so by pressuring the cooking space. Its not quite salt, but you could find one in a kitchen.

Commonly set to HIGH or LOW using a weight on the vent Setting LOW would be 0.4~0.55 bar and HIGH would be 0.9~1 bar so double atmospheric pressure. Readings are on top of the 1 bar normal atmospheric pressure of the room.

By comparison, a car tyre would be about 2.2 bar, so just over triple normal room pressure air.

0.5 bar would be a boiling point of 112 degrees C

1.0 bar would boil at 120 degrees C

2.6 bar would be 140 degrees C, but good luck getting such pressures in the kitchen.

Disclaimer, even 0.5 bar is a lot of pressure, and any rupture will cause significant damage.

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  • 2
    $\begingroup$ Maybe you should mention that these values refer to gauge pressure, and not to absolute pressure which is normally used in chemistry. $\endgroup$
    – user7951
    Jun 3 '19 at 17:03

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