The sodium salt of a carboxylic acid `(A)` was produced by passing a gas `(B)` into an aqueous solution of caustic alkali at an envolved temperature and pressure `(A)` on heating in the presence of sodium hydroxide followed by the treatment with sulphuric acid gave a dibasic acid `(C)`. A sample of `0.4 gm` of acid `(C)` on combustion gave `0.08 gm` of water,`.39gm` of `CO_2` and weighting `1.0 gm` on ignition yielded `0.71 gm` of silver as residue. Identify `(A), (B),` and (C)`.

Explanation

Calculation of molecular weight of the acid (C ): As the acid is dibasic, the Ag salt of it will contain 2 atoms of Ag in one molecule of the acid
`{:("Acid",rarr,"Salt containing",rarr,Ag),((C),,"2 Ag atoms",,0.71g),(,,1g,,):}`
Applying POAC for Ag atom in the second step, `2 xx`moles of the salt= moles of Ag in the product
`2 xx (1)/(M)= (0.71)/(108) ((M -="mol wt"),("of salt"))`
`therefore M ~~ 304`
Hence, mol wt of the acid `=(304 -2 xx 108+2)`= 90
Calculation of empirical formula weight and molecular formula of the acid (C ). Moles of `C= (0.39)/(44)= 0.0088`
Wt of `C= 0.0088 xx 12= 0.1064g`
Moles of `H= (2 xx 0.08)/(18)= 0.0088`
Wt. of `H= 0.0088 xx 1= 0.0088g`
`therefore` wt of `O= [0.4 -(0.1064 + 0.0088)]=0.2848`
Moles of `O= (0.2848)/(16)=0.0178`
`therefore` moles of `C: H: O= 0.0088: 0.0088: 0.0178= 1 : 1: 2`
`therefore` empirical formula is `CHO_(2)` (45)
As the mol. wt is 90, molecular formula is `(COOH)_(2)`
As the acid (C ) is now known, the reaction sequence may be represented as
`{:(NaOH + CO ("gas"),rarr,HCOONa,rarr,(COOH)_(2)),(" "(B),," "uarr,," "(C )),(,,HCOOH,,),(,," "(A),,):}`