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2 g of benzoic acid (C(6)H(5)COOH) disso...

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 ?

Text Solution

Verified by Experts

The given quantities are :
`w_(2)=2"g, K"_("f")=4.9" K kg mol"^(-1),w_(1)=25" g"`
`DeltaT_("f")=1.62" K"`
Substituting these values in equation, we get :
`M_(2)=(K_("f")xx w_(2)xx1000)/(DeltaT_("f")xx w_(1))`
`M_(2)=(4.9" K kg mol"^(-1)xx2g xx1000" g kg"^(-1))/(25" g"xx1.62" K")=241.98" g mol"^(-1)`
Thus, experimental molar mass of benzoic in benzene is `=241.98" g mol"^(-1)`
Now consider the following equilibrium for the acid :
`2" C"_(6)"H"_(5)"COOH"iff(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 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=2xx0.496=0.992`
Therefore, degree of association of benzoic acid in benzene is `99.2%`.
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