Discussion Forum

1)

The plot given below shows p-T curves (where p is the pressure and T is the temperature ) for two solvents X and Y and isolmolal solution of NaCl  these solvents. NaCl completely dissociates in both the solvents

On the addition of the equal number of moles of a non -volatile S in equal amount (in kg) of these solvents, the elevation of boiling point X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is........... 

Answer:

Option A

Explanation:

From the graph we can note $\triangle T_{b}$ for solution X i.e.,

 $\triangle T_{b(X)}=362-360=2$

 Like wiise, $\triangle T_{b}$ for solution Y i.e.,

 $\triangle T_{b(Y)}=368-367=1$

Now by using the formula

    $\triangle T_{b}$ =i × molality of solution × K_b

For solution X

       2= i × m_NaCl × K_b(X)          ...............(i)

Similarly for solution Y 

 1= i × m_Nacl × Kb(Y)            ..................(ii)

from Eq .(i) and (ii) above

          $\frac{K_{b(X)}}{K_{b(Y)}}=\frac{2}{1}=2$

or    K_b(X) = Kb(Y)

  For solute S 

           $2S\rightarrow S_{2}$    (Given due to dimerisation)

initial         α                 0

final         (i - α )           α/2

So, here     $i=[1-\frac{\alpha}{2}]$

      $\triangle T_{b[X](s)}= (1-\frac{\alpha_{1}}{2})K_{b(X)}$

     $\triangle T_{b[Y](s)}= (1-\frac{\alpha_{2}}{2})K_{b(Y)}$

 Given    $\triangle T_{b(X)(s)}=3\triangle T_{b(Y)(s)}$

    $(1-\frac{\alpha_{1}}{2})K_{b(X)}= 3 \times (1-\frac{\alpha_{2}}{2}) \times K_{b(Y)}$

 or    $2(1-\frac{\alpha_{1}}{2})= 3  (1-\frac{\alpha_{2}}{2})$   [ $\because  K_{b(X)}=2K_{b(Y)}$ ]

  or           $2(1-\frac{\alpha_{1}}{2})= 3  (1-\frac{0.7_{}}{2})$   [ as given,α₂ =0.7 ]

  or         4- α₁ =6 -2.1          or  2 α₁ =0.1

            so,      α  =0.05