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10 g of substance below were dissolve in each 1 L water:

$\ce{Na2SO4}$
$\ce{NaCl}$
Glucose
$\ce{MgCl2}$

Which one has the highest boiling point?

Glucose has London dispersion force. $\ce{Na2SO4},$ $\ce{NaCl},$ $\ce{MgCl2}$ have ionic forces. Ionic forces is stronger than London dispersion force, so $\ce{Na2SO4},$ $\ce{NaCl},$ $\ce{MgCl2}$ have higher b.p. than glucose.

Now, how to decide which one has highest b.p. among $\ce{Na2SO4},$ $\ce{NaCl},$ $\ce{MgCl2}?$

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  • $\begingroup$ Tech note: in chemistry "BP" is a poor choice for an abbreviation for boiling point as it's ambiguous and may also mean boron phosphide. A more or less standardized abbreviation would be "b.p." $\endgroup$
    – andselisk ♦
    Jul 19, 2019 at 7:57
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    $\begingroup$ The water boils, not the solutes. All the ionic interactions are gone in water, replaced by ion dipole. Solid glucose makes hydrogen bonds with itself, replaced by hydrogen bonds with water in solution. $\endgroup$
    – Karsten Theis
    Jul 19, 2019 at 14:25

2 Answers 2

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Boiling point elevations proportional to the molar concentration of dissolved particles. So that's what you have to figure out, being careful to (1) properly convert mass to moles and (2) recognizing that some of the solutes can form ions so you can get more than one particle per "molecule".

Let's look at glucose. If you work out the molecular mass it's about 180 grams per mole. Divide that into 10 grams and you get about 0.056 moles of glucose molecules in your one liter of solution.

Now try sodium sulfate. For this one you have a "molecular" mass of 142 grams per mole, so you seem to get about 0.070 mole of particles. Except ... Sodium sulfate forms ions, each "molecule" gives one sulfate ion and two sodium ions. Thus three particles overall, and you have to multiply your 0.070 figure by that factor of 3. Thus the sodium sulfate solution will have 0.21 mole of ions in one liter. The sodium sulfate solution will therefore give a higher boiling point than the glucose solution.

Work it out now with the other two choices and pick the one with the most particles in solution including ion formation. Be sure to use reasonably accurate atomic masses because this set of choices could throw you for a loop.

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  • $\begingroup$ The answer is Na2SO4 right? But the key answer is MgCl2 $\endgroup$
    – Lifeforbetter
    Jul 22, 2019 at 14:55
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    $\begingroup$ Actually I get sodium chloride. Not sure where the key gets magnesium chloride, but if you work out the numbers you should have found that both chlorides beat sodium sulfate. $\endgroup$
    – Oscar Lanzi
    Jul 22, 2019 at 17:35
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Normally, I would not respond to this kind of question since it is most likely a homework but I am considering this question as an exception.

One important equation to determine the boiling point of ionic solutions is the boiling point elevation equation which states that the change in boiling temperature of the pure solvent is equal to imKb; where i is the vant hoff factor, m is the molality of the solution, and Kb is the ebullioscopic constant of the solvent.

Assuming that the solution is present in equimolal concentrations and in the same solvent, the arrangement of boiling points would be:

Na2SO4 = MgCl2 > NaCl

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  • $\begingroup$ Thanks but the answer is MgCl2 why? $\endgroup$
    – Lifeforbetter
    Jul 19, 2019 at 7:56
  • $\begingroup$ the vant hoff factor for the solution is equal to the number of species dissolved in the species. NaC has only a vant hoff factor of 2 since the dissolved species in a brine solution is Na+ and Cl-. However, Na2SO4 and MgCl2 each has a vant hoff factor of 3. $\endgroup$
    – Kent de los Reyes
    Jul 19, 2019 at 7:59
  • $\begingroup$ @KentdelosReyes It seems like this should be added to the answer alongside with the data source. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – andselisk ♦
    Jul 19, 2019 at 8:00
  • $\begingroup$ Correct also the consideration made by OP on dispersion forces versus ionic bonds which are not (relevant (if not indirectly) here. $\endgroup$
    – Alchimista
    Jul 19, 2019 at 10:33
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    $\begingroup$ I don't know.. @oscar $\endgroup$
    – Lifeforbetter
    Jul 22, 2019 at 17:25

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