--%>

What is Elevation in boiling point?

The boiling of a liquid may be defused by the temperature at which its vapour pressure which is equal to atmospheric pressure. The effect of addition in a non-volatile solute on the boiling point shown and its solution containing non-volatile solute with temperature are represented by the curves AB and CD respectively. It is evident by the curves temperature the vapour pressure of solutions is lower than that of the pure solvent and thus, the vapour pressure.

Curve for solution runs below that of the pure solvent. At temperature T0, the vapour pressure of the pure solvent becomes equal to the atmospheric pressure. Thus, the atmospheric pressure and therefore, it is necessary to heat the solution to a higher temperature sayT1 in the atmospheric pressure. Thus, it is clear that the solution in higher temperature than the pure solvent. Evidence T1 - T0 (or Δ Tb)is the elevation in boiling point vapour pressure (Δp), the elevation in the boiling point is also proportional to the solute concentration. Thus,

ΔTb ∝ Δp

According to Raoult's law, Δp ∝ xB

∴ ΔTb ∝ xB

1444_elevation in boiling point.png 

1084_elevation in boiling point1.png 

1730_elevation in boiling point2.png 

1333_elevation in boiling point3.png 

If WA is the mass of solvent in kg, then  1283_elevation in boiling point4.png is equal to molality (m) of the solution

ΔTb = kMA m

Here, k and MA are constants and hence their product, i.e. kMA is replaced by another constant K2.

ΔTb = Kb m, where Kb is called boiling point-elevation constant or molal elevation constant or molal ebullioscopic constant.

As elevation in boiling point depends upon the relative number of moles of solute and solvent but does not depend upon the nature of solute, so it is a colligative property.

   Related Questions in Chemistry

  • Q : Diffusion Molecular View When the

    When the diffusion process is treated as the movement of particles through a solvent the diffusion coefficient can be related to the effective size of diffusing particles and the viscosity of the medium.To see how the experimental coefficients can be treat

  • Q : Examples of reversible reaction

    Describe some examples of a reversible reaction?

  • Q : Concentration of urea Help me to go

    Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<

  • Q : Dependcy of colligative properties

    Colligative properties of a solution depends upon: (a) Nature of both solvent and solute (b) The relative number of solute and solvent particles (c) Nature of solute only (d) Nature of solvent only

  • Q : Mole fraction and Molality Select the

    Select the right answer of the following question.What does not change on changing temperature : (a) Mole fraction (b) Normality (c) Molality (d) None of these

  • Q : Precipitation problem On passing H 2 S 

    On passing H2S  gas through a solution of Cu+ and Zn+2 ions, CuS is precipitated first because: (i) Solubility product of CuS is equal to the ionic product of ZnS (ii) Solubility product of CuS is equal to the solubility product o

  • Q : Advantages of doing your own chemistry

    What are the advantages of doing your own chemistry assignments? State your comment?

  • Q : Molecular mass from Raoults law Provide

    Provide solution of this question. Determination of correct molecular mass from Raoult's law is applicable to: (a) An electrolyte in solution (b) A non-electrolyte in a dilute solution (c) A non-electrolyte in a concentrated solution (d) An electrolyte in a liquid so

  • Q : Problem on moles of solution The number

    The number of moles of a solute in its solution is 20 and total no. of moles are 80. The mole fraction of solute wil be: (a) 2.5 (b) 0.25 (c) 1 (d) 0.75

  • Q : Determining concentration in ppm A 500

    A 500 gm tooth paste sample has 0.2g fluoride concentration. Determine the concentration of F in terms of ppm level: (a) 250 (b) 200 (c) 400 (d) 1000Answer: (c) F-ions in ppm = (0.2/500) x 106 = 400