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Boiling Point Calculator
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Boiling point is defined as the temperature at which a liquid and the vapor above it are in equilibrium. At normal atmospheric pressure, the boiling point of a liquid will be the temperature at which the liquid is at equilibrium with the atmosphere above the liquid. Boiling points of solutions are higher that the boiling points of the pure solvents.

Another way to say this is that boiling point is the temperature at which molecules leave the liquid phase and are moved into gas phase.

The formula for boiling point elevation is given below.

$\Delta$ T = Kbm

The boiling point of the solution is given by

TSolution = TPure solvent + $\Delta$ T
 

Steps for Boiling Point Calculator

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Step 1 :  

Read the problem and list out the values given.



Step 2 :  

Substitute the values in the formula to find out the unknown value.


$\Delta$ T = Kbm
TSolution = TPure solvent + $\Delta$ T



Problems on Boiling Point Calculator

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  1. If 10.0g of NaCl is added to 6.00L of water and that the density of the solution is 1.00g/ml. Also assume that this very dilute solution behaes ideally. The value of Kb for water is 0.512oC/m


    Step 1 :  

    Given data


    Kb = 0.512oC/m


    10.0g NaCl $\times \frac{1\ mol\ NaCl}{58.5\ g\ NaCl}$ = 0.171 mo NaCl
    6.00 L water $\times \frac{1.00\ kg\ water}{L\ water}$ = 6.00 kg water
    $\frac{0.171\ mol\ NaCl}{6.00\ kg\ water}$ = 0.0285 m NaCl



    Step 2 :  

    Substitute the values in the formula.


    $\Delta$ T = Kbm


    $\Delta$ T = 0.512 $\times$ 0.0285m


    $\Delta$ T = 0.03oC



    Answer  :  

    The answer if $\Delta$ T = 0.014oC



  2. A solution of 10.0g of sodium chloride is added to 100.0g of water in an attempt to elevate the boiling point. What is the boiling point of the solution?


    Step 1 :  

    Given data


    Moles of NaCl = $\frac{10.0}{58.5}$ = 0.171 mol
    Molarity = $\frac{0.171}{0.100}$ = 1.71 m



    Step 2 :  

    For NaCl
    Kb(water) = 0.52oC/m
    $\Delta$ Tb = Kbm
    $\Delta$ Tb = (2)(0.52)(1.71)
    $\Delta$ Tb = 1.78oC


    Tsolution = Tb(pure solvent) + $\Delta$Tb


    Tsolution = 100oC + 1.78


    Tsolution = 101.78oC



    Answer  :  

    The answer is Tsolution = 101.78oC



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