What is the boiling point of an aqueous solution that has a vapor pressure of $23.0~\text{Torr}$ at $25~^\circ\text{C}$?

$(P^\circ_{\ce{H2O}}=23.78~\text{Torr};~K_\mathrm b= 0.512~^\circ\mathrm{C\ kg/mol})$

To solve this I used the equation $\Delta T_\mathrm b =m\times K_\mathrm b$. I calculated a molality of $1.88247\ \mathrm{mol/kg}$ which I then used to calculate a boiling point of $-0.9638\ \mathrm{^\circ C}$. This answer is apparently wrong and I don't understand why.


1 Answer 1


If molality is 1.88, then the change in boiling point is -0.96 degrees C, not the boiling point itself.

  • $\begingroup$ But if i subtract the change in temperature from the normal boiling point of water (0 C) then we get a boiling point of -0.96 C, which is wrong. $\endgroup$
    – Nate
    Jan 28, 2015 at 1:48
  • 5
    $\begingroup$ Water boils at 100 degrees C $\endgroup$
    – DavePhD
    Jan 28, 2015 at 2:19

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