2 g of benzoic acid `(C_(6)H_(5)COOH)` dissolved in 25g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg `"mol"^(-1)`. What is the precentage association of acid if it forms dimer in solution ?

Explanation

The given quantities are : `omega_(2) = 2 g , K_(f) = 4.9 " K kg mol"^(-1) , omega_(1) = 25 g `
`Å T_(f) =1 .62 K `
Substituting these values in equation (2.36) we get:
`M_(2) = (4.9 " kg mol"^(-1) xx 2 g xx 1000" g kg"^(-1))/(25 g xx 1.62 K ) = 241 . 98 " g mol"^(-1)`
Thus, experimental molar mass of benzoic acid in benzene is
` = 241 . 98 " g mol"^(-1)`
Now consider the following equilibrium for the acid:
` 2 C_(6)H_(5) COOH hArr (C_(6)H_(5)COOH)_(2)`
If x represents the degree of association of the solute then we would have (1 – x ) mol of benzoic acid left in unassociated form and correspondingly `x/2` as associated moles of benzoic acid at equilibrium Therefore, total number of moles of particles at equilibrium is:
` 1 - x + x/2 = 1 - x/2 `
Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i.
But i = `("Normal molar mass ")/( "Abnormal molar mass ")`
` = (122 "g mol"^(-1))/(241 . 98 "g mol"^(-1))`
or `x/2 = 1 - 122/(241.98) = 1 - 0.504 = 0.496`
or ` x = 2 xx 0.496 = 0.992 `
Therefore, degree of association of benzoic acid in benzene is 99.2 %.