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 isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.

On addition of equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent 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 ______________.

Explanation

Case: 1
For solvent X
Solute: NaCl
Molality : `m_(1)`
`i=2`
`DeltaT_(b)=(362-360)=2` (from graph)
`DeltaT_(b)=i(m)(k_(b))_(x)`
`2=(2)(m)(K_(B))_(x)`……….1 , `(K_(b))_(x)=1/(m_(1))`
For solvent Y
Solute : NaCl
MOlality : `m_(2)`
`i=2`
`DeltaT_(b)=(368-367)=1` (from graph)
`DeltaT_(b)=im(K_(b))_(y)`
`1=(2)(m_(1))(K_(B))^(y)` ......2 , `(K_(b))_(y)-1/(2m_(2))`
Dividing 1 by 2
`2/1=(2m_(1)(K_(b))_(x))/(2m_(2)(k_(b))_(y))`
`((K_(b))_(x))/((K_(b))_(y))=2`...........3
Case : 2
For solvent X:
Solute: Non volatile
Molality : `m_(2)`
`i=i_(x)`
`DeltaT_(b)=3a`
`DeltaT_(b)=i_(x)m(K_(b))_(x)`
`3a=i_(x)m_(2)(K_(b))_(x)`........4
For solvent Y:
Solute: Non volatile
Molality : `m_(2)`
`i=i_(y)`
`DeltaT_(b)=a`
`DeltaT_(b)=i_(y)m(K_(b))_(y)`
`a=i_(y)m_(2)(K_(b))_(y)` .....5
Dividing 4 by 5
`(3a)/a=(i_(x))/(i_(y)) m/n ((K_(b))_(x))/((K_(b))_(y))" "," "(i_(x))/(i_(y))=3xx((K_(b))_(y))/((K_(b))_(x))=3/2`
Now `(i_(x))/(i_(y))=3/2`
In solvent Y:
`underset(1)(2S)tounderset(1)(S_(2))`
`(1-0.7)0.7//2,i_(y)=1-0.7/2=(2-0.7)/2=1.3/2`
In solvent X :
`underset(1)(2S)tounderset(1)(S_(2))`
`1-alpha" "(alpha)/2" ", i_(x)=1-(alpha)/2`
`(i_(x))/(i_(y))=3/2=((1-0.5alpha))/1.3xx2`
`3/2=((1-0.5alpha))/1.3xx2" " , " "(3xx1.3)/4=(1-0.5alpha)`
`alpha=0.05`
Higher the no. of particles, higher is the elevation in B.P.