Question

At hundred degree celsius the vapour pressure of a solution of 6.5 gram of solute in hundred gram of water is 732 mm is kps 0.52 the boiling point of the solution will be

Answer 1

Answer:101°C

Explanation:

We know that relative lowering of vapour pressure is given by the formula

 

                                               P0A− PAP0A = Xsolute

                 P0A of pure water =  760 mm Hg

                  PA   of the solution  = 732 mm Hg

                                       This implies  Xsolute  = 760−732760   = 0.037

 Now as we have got the mole fraction of the solute , we can find the no. of moles of solute present

                                                0.037 =  nsolutensolute +  10018

                    This gives ,    nsolute    =    0.213 

      Hence we have 0.213 moles of the solute present . Now we know the formula for elevation in boiling point.

                                ΔTb       =  kb * m     = 0.52 * 0.2131001000 = 1.1076

                              Hence    ΔTb  =  1.1076

Therefore the boiling point will be = (100 + 1.1076) =  101.1076o C

     Hence the boiling point of the solution will be 101.1076o C.