1.0 g sample of the Rochelle salt [NaOOC-CHOH-CHOH-COOK] (NaKC_(4)H_(4)O_(6).4H_(2)O)(Mw=282), on ignition, is converted into NaKCO_(3)(Mw=122), which is titrated with 50 " mL of " 0.1 MH_(2)SO_(4). The excess of H_(2)SO_(4) requries 30 mL 0.2 M KOH.

1.0 g sample of the Rochelle salt [NaOOC-CHOH-CHOH-COOK] (NaKC_(4)H_

1.	1.0 g sample of the Rochelle salt [NaOOC-CHOH-CHOH-COOK] (NaKC_(4)H_(4)O_(6).4H_(2)O)(Mw=282), on ignition, is converted into NaKCO_(3)(Mw=122), which is titrated with 50 " mL of " 0.1 MH_(2)SO_(4). The excess of H_(2)SO_(4) requries 30 mL 0.2 M KOH.
Answer» Solution :`NaKC_(4)H_(4)O_(6).4H_(2)Ooverset(Delta)toNaK.CO_(3)+3CO_(2)uarr+H_(2)Ouarr` Total m" Eq of "`H_(2)SO_(4)=50xx0.1xx2` (n factor) `=10.0` EXCESS m" Eq of "`H_(2)SO_(4)=m" Eq of "NAOH used` `=30xx0.2xx1` (n factor) `=6.0` m" Eq of "`H_(2)SO_(4)` used for `NaKCO_(3)` `=m" Eq of "H_(2)SO_(4)` (total)`-m" Eq of "H_(2)SO_(4)` used for NaOH `=10-6=4.0` `thereforem" Eq of "NaKCO_(3)=4.0` `implies` n-factor of `NaKCO_(3)=2`, during its neutralisation with `H_(2)SO_(4)(" Eq of "NaKCO_(3)=(122)/(2))` Weight of `NaKCO_(3)=4.0xx10^(-3)XX(122)/(2)=0.244g` 1 " MOL of "Rochelle salt`=1 " mol of "NaKCO_(3)` Therefore, weight of rochelle salt`=(282xx0.244)/(122)=0.364g` `%` purity of Rochelle salt`=(0.364xx100)/(1.0)=36.4%`