# Boiling Point Help?

My teacher recently gave us a worksheet and I'm currently stuck on the last problem.

'Assume you have 7.21 m solution that raised the boiling point of the solution by 1.25°C. What is the molal boiling point constant (Kb) of the solution?'How do I solve for this? I already know how to solve using Delta T=KbM for reference, am I just suppose to interchange the formula?

### 2 Answers

- Dr WLv 71 month ago
actually.. you can't solve for Kb without more information. from Blagdon's law

.. dTbp = Kb * m * i

where

.. dTbp = change in bp = bp solution - bp pure solvent = 1.25°C (in this case)

.. Kb = ebullioscopic constant (aka boiling point elevation constant).

.. m = molality = moles solute / kg solvent = 7.21m in this case

.. i = van't hoff factor = # particles 1 particle of solvent dissociated into in solution

examples of i

.. 1 NaCl -> 1 Na+ + 1 Cl-.... 1 particle --> 2 particles... i = 2

.. 1 MgCl2 --> 1 Mg(2+) + 2 Cl-... 1 particle --> 3 particles.. i=3

.. 1 glucose --> 1 glucose.... i = 1

***********

in this problem, you have no way of knowing "i". So the best you can do is this

.. dTbp = Kb * m * i

.. Kb = dTbp / (m * i)

.. Kb = 1.25°C / (7.21m * i)

if you make the assumption that i = 1,

. Kb = 1.25°C . 7.21M = 0.173 °C/m

but I wouldn't. I'd verify the problem statement and ask for guidance from your instructor about the van't hoff factor.

- BobbyLv 71 month ago
Delta T=Kbm

make Kb the subject of the formula

Kb =1.25oC / 7.21m = 0.173 °C m^-1