Two grams of benzoic acid `(C_(6)H_(5)COOH)` dissolved in `25.0 g` of benzene shows a depression in freezing point equal to `1.62 K`. Molal depression constant for benzene is `4.9 K kg^(-1)"mol^-1`. What is the percentage association of acid if it forms dimer in solution?

Explanation

Step 1. Calculate van't hoff factor
`DeltaT_(f)=iK_(f)m`
or `i=(DeltaT_(f))/(K_(f)m)`
`=((1.62 K))/((4.9 K kg mol^(-1))((2g)/(122 g mol^(-1)xx0.025 Kg)))`
`=0.504`
Alternatively, we can determine van't hoff factor as follows:
`i=((DeltaT_(f))_("observed"))/((DeltaT_(f))_("calculated"))`
`(DeltaT_(f))_("observed") =1.62 K`
`(DeltaT_(f))_("calculated")=K_(f)m`
`=(4.9 K kg mol^(-1))((2 g)/(122 g mol^(-1)xx0.025 kg))`
`=3.21 K`
Therefore
`i=(1.62 K)/(3.21 K)=0.504`
We can also go like this
`i=("calculated molar mass")/("observed molar mass")`
Calculated molar mass=`(12xx7+6xx1+16xx2)=122 g mol^(-1)`
Observed molar mass=`(K_(f) w_(solute))/(DeltaT_(f) w_(solvent))`
`=((4.9 K kg mol^(-1))(2 g))/((1.62 K)(0.025 kg))`
`=241.98 g mol^(-1)`
Therfore
`i=(122 g mol^(-1))/(241.98 g mol^(-1))`
`=0.504`
Step 2: Determining degree of dissociation
`2C_(6)H_(5)CO_(2)H hArr (C_(6)H_(5)COOH)_(2)`
Moles before assocation 1 mol 0 mol
Mole After assocation `(1 - alpha)` mol `(alpha)/(2)` mol
`i=("Moles after association")/("Moles before association")`
`=(1- alpha+ alpha//2)/1=(1- alpha//2)/1=1-alpha//2`
or `alpha=(1-i)xx2`
`=(1-0.504)(2)`
`=(0.496)(2)`
`=0.992`
Therefore
percentage association of benzoic acid `=0.992xx100%=99.2%`