When `1.22 g C_(6)H_(5)COOH` is added into two solvents, the following data of `DeltaT_(b)` and `K_(b)` are obtained:
i. In `100 g CH_(3)COCH_(3)`,`DeltaT_(b)=0.17`,`K_(b)=1.7 kg K mol^(-1)`.
ii. In `100 g` benzene,`DeltaT_(b)=0.13` and `K_(b)=2.6 kg K mol^(-1)`.
Find out the molecular weight of `C_(6)H_(5)COOH` in both cases and interpret the results.

Answer

`Mw_(2)=244`

Explanation

`(i) In first case
By `DeltaT_(b)=K_(b)xxm`
(Given,` DeltaT_(b)=0.17`,`K_(b)=1.7 kg K "mol"^(-1)`)
`0.17=1.7 xx 1.22/(Mwxx100xx10^(-3))`
So, `Mw=122`
`(ii) In second case
(Given,` DeltaT_(b)=0.13`,`K_(b)=2.6 kg K "mol"^(-1)`)
`0.13=2.6 xx 1.22/(Mwxx100xx10^(-3))`
So, `Mw=244`
The conclusion is that benzoic acid dimerizes in benzene as
`HA+NaOHhArrNa^o+A^(ө)+H_2O`
b. Thus, at the end point, the solution contains only `Na` `A`(being salt of weak acid and strong base) whose concentration is `0.05 M`.
`A^(ө)+H_2OhArrHA+OH^(ө)`
`sqrt((10^(-14)xx5.6xx10^(-6))/0.05)=10^(-9)xxsqrt(1.12)`
So, `pH=9-1/2 log 1.12=8.975`