Neutralisation of `0.3504 gm` of acid `(A)` requires `27.24 ml` of `0.15 M NaOH`, and the `Mw` is found from the mass spectral data to be `172.1 gm mol^-1`.
(a) Calculate the `N.R` of `(A)`.
(b) How many ionisable `H` atom are there in `(A)` ?

Explanation

(a) `mEq. Of NaOH = 27.24 xx 0.15 = 4.086 mEq`.
=`(4.086)/(1000) eq. = 0.004086 eq.`
`N.E (Ew) = (0.3504)/(0.004086) = 85.76 gm eq^-1`.
(b) Ionisable `H` atoms `= (mw)/(N.E)` or `(Mw)/(Ew)`.
=`(172.1)/(85.76) = 2 eq. mol^-1`
=`2` ionisable `H` atoms `mol^-1`
=`A` is dicarboxylic acid.