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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» <html><body><p></p>Solution :`NaKC_(4)H_(4)O_(6).4H_(2)Ooverset(Delta)toNaK.CO_(3)+3CO_(2)uarr+H_(2)Ouarr` <br/> Total m" Eq of "`H_(2)SO_(4)=50xx0.1xx2` (n factor) `=10.0` <br/> <a href="https://interviewquestions.tuteehub.com/tag/excess-978535" style="font-weight:bold;" target="_blank" title="Click to know more about EXCESS">EXCESS</a> m" Eq of "`H_(2)SO_(4)=m" Eq of "<a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> used` <br/> `=30xx0.2xx1` (n factor) <br/> `=6.0` <br/> m" Eq of "`H_(2)SO_(4)` used for `NaKCO_(3)` <br/> `=m" Eq of "H_(2)SO_(4)` (total)`-m" Eq of "H_(2)SO_(4)` used for NaOH <br/> `=10-6=4.0` <br/> `thereforem" Eq of "NaKCO_(3)=4.0` <br/> `implies` n-factor of `NaKCO_(3)=2`, during its neutralisation with <br/> `H_(2)SO_(4)(" Eq of "NaKCO_(3)=(<a href="https://interviewquestions.tuteehub.com/tag/122-271030" style="font-weight:bold;" target="_blank" title="Click to know more about 122">122</a>)/(2))` <br/> Weight of `NaKCO_(3)=4.0xx10^(-3)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(122)/(2)=0.244g` <br/> 1 " <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> of "Rochelle salt`=1 " mol of "NaKCO_(3)` <br/> Therefore, weight of rochelle salt`=(282xx0.244)/(122)=0.364g` <br/> `%` purity of Rochelle salt`=(0.364xx100)/(1.0)=36.4%`</body></html>


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