At 100`C the vapour pressure of a solution of 6.5g of a solute in 100g water is 732mm.if Kb=0.52,the boiling point of this solution will be
 

Dear Student ,
                      We know that relative lowering of vapour pressure is given by the formula
 
                                               $\frac{{P^0}_A- P_A}{{P^0}_A} = X_{solute}$

                 P⁰_A of pure water =  760 mm Hg
                  P_A   of the solution  = 732 mm Hg

                                       This implies  $X_{solute } = \frac{760-732}{760}$   = 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 =  $\frac{n_{solute}}{n_{solute} + {\displaystyle \frac{ 100}{18}}}$

                    This gives ,    n_solute    =    0.213 

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

                                ΔT_b       =  k_b * m     = $\frac{0.52 * 0.213}{{\displaystyle \frac{100}{1000}}} = 1.1076$

                              Hence    ΔT_b  =  1.1076

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

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

I hope you understood . Keep posting.