Correct Answer - B
E.F.Mass =12+1+32=45
VD of ester =73
Mol. Mass of ester `=2xx73=146`
Mol. Mass of acid =Mol. Mass of ester
`-2 xx` Mol, Mass of `C_(2)H_(5)+2xx At.` mass of H
`=146-2xx29+2`
`=146-58+2=90`
`n=("Mol.mass" )/("EF mass") or n =90/45=2 `
`therefore "Mol Formula" = 2xxCHO_(2)=C_(2)H_(2)O_(4).`