Question

Calculate the boiling point elevation for a solution prepared by adding 10 g of CaCl2 to 200 g of water.
(Kb for water = 0.512 K kg mol-1, Molar mass of CaCl2 = 111 g mol-1)

Answer 1

Elevation in Boiling Point is given By
∆Tb = Kb m                       ...... eq 1

where ∆Tb = Elevation in Boiling Point

Kb= molal elevation constantm = molality of the solution

As we know

 m=w2/M2×1000/w1              .......eq 2

where w2 is mass of solute(in grams)          
M2 is mass of solute(in grams)          

w1 is mass of solute(in grams)


According to the Question 
Given: mass of CaCl2 (w2) = 10 g
            mass of water (w1 ) = 200 g 
            Molar mass of CaCl2 (M2) = 111 g mol​-1
            Molal Elevation Constant = 0.512 K kg mol-1

Substituting the values in the equation 2
m= 10111×1000200   = 0.450 mSubstituting the value of m in equation 1
∆Tb = Kbm       
= 0.512×0.450       
  = 0.2304 K 
 

Answer 2

Answer:

The elevation in the boiling point for solution of 10g of CaCl₂ to 200g of water is 0.2306K.

Explanation:

Elevation in boiling point: Δ$T_{b}=K_{b}m$         ..........(1)

where m is molality and  $K_{b}$ is molal elevation constant  

Molality m= (moles of solute /mass of solvent) ×1000

Mass of CaCl₂ $=10g$

Molar mass of CaCl₂  $= 111gmol^{-1}$

mass of water(solvent)  $= 200g$

moles of CaCl₂  $=\frac{10}{111} =0.09moles$

Molality  $m=\frac{0.09}{200}*1000 =0.450molKg^{-1}$

Given $K_{b} =0.512Kkgmol^{-1}$

Now     Δ$T_{b}=(0.512 K kg mol^{-1})(0.450molkg^{-1})$

              Δ$T_{b}=0.2306K$

Therefore  the boiling point elevation for a solution  is 0.2306K.