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

Relevance
  • Dr W
    Lv 7
    1 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. 

  • Bobby
    Lv 7
    1 month ago

    Delta T=Kbm

    make Kb the subject of the formula 

    Kb =1.25oC / 7.21m = 0.173 °C m^-1

      

Still have questions? Get answers by asking now.