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

On addition of equal number of moles of nonvolatile solute S in equal amount ( in kg ) of these solvents, the elevation of boiling point of solvent X is three times that solvent Y. Solute S is known to undergo dimerization is 0.7 in solvent Y , then what is the degree of dimerization in solvent X ?

Explanation

When NaCl is solute ,
` i= 2`
`Delta T_(b) = ixxk_(b) xx m`
For solvent X
`Delta T_(b) = 362 - 360 = 2K ` (from graph)
`2 = 2xx (k_(b))_(X) m" "…..(i)`
For solvent Y .
`Delta T_(b) = 368 - 367 = 1 K` (from graph) .
` 1 = 2 xx (k_(b))_(Y) m" "......(ii)`
Dividing (i) by (ii)
`((k_(b))X)/((k_(b))Y) = 2`

Now for solute , S which dimerizes in both solvent
`2S hArr S_(2)`
`{:(,"Initial",1,0),(,"After association",1-alpha,alpha//2):}`
Total no. of moles ` = 1- alpha + alpha//2 =1 - alpha//2`
` i = 1 -alpha//2`
`(DeltaT_(b)_(X) = (1-alpha//2) (K_(b))_(X) xx m" "........(iii)`
`(Delta T_(b))_(Y) = (1-(0.7)/(2)) (k_(b))_(Y) xx m = 0.65 (k_(b))_(Y) xx m" ".....(iv)`
Elevation in boiling point in solvent X and Y
`((Delta T_(b))_(X))/((Delta T_(B))_(Y)) = 3` (Given).
Dividing eq.(iii) by (iv) for same concentration.
`3 = (1+alpha//2)/(0.65) xx ((k_(b))_(X))/((k_(b))_(y)) `
`3 = (1-alpha//2)/(0.65) xx 2`
` 1- alpha //2 = (3xx 0.65)/(2) = 0.975`
` -alpha//2 = 0.9751-1 = - 0.025`
or `alpha = 0.05`