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1 g of monobasic acid in 100 g of water ...

`1 g` of monobasic acid in `100 g` of water lowers the freezing point by `0.168^(@)`. If `0.2 g` of same acid requires `15.1 mL mol^(-1)` of `N//10` alkali for complete neutralization, calculate the degree of dissociation of acid. `K'_(f)` for `H_(2)O` is `1.86 K mol^(-1) kg`.

Text Solution

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`DeltaT = (1000xxK_(f)xxW_(2))/(Mw_(2)xxW_(1))`
and `DeltaT = 0.168 , W_(2)=1 g , W_(1)=100 g` and `K'_(f)=1.86`
`:. Mw_("acid")=110.71` (This is exp.mol.wt.)
Now, mEq of acid = mEq of alkali
`(0.2)/(E )xx1000 = 15.1xx(1)/(10)`
`:.` Ew of acid `= (0.2xx1000xx10)/(15.1)=132.42`
`:.` Normal Mw `=132.45xx1`
(Since monobasic)
`{:(,HA,hArr,H^(o+),+,A^(Θ)),("Before dissociation",1,,0,,0),("After dissociation",(1-alpha),,alpha,,alpha):}`
Now, `(Mw_("Normal"))/(Mw_(Exp))=(132.45)/(110.71)=1+alpha implies alpha =19.6 %`
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