A 0.518 g sample of lime stone is dissolved in HCl and then the calcium is precipitated as CaC_2O_4 After filtering and washing the precipitate, it requires 40.0 mL of 0.250 N KMnO_4solution acidified with H_2SO_4to titrate it as, MnO_(4)^(-)+H^(+) +C_(2)O_(4)^(2-) rarr Mn^(2+)+CO_(2)+2H_(2)O The percentage of CaO in the sample is:

A 0.518 g sample of lime stone is dissolved in HCl and then the calciu

1.	A 0.518 g sample of lime stone is dissolved in HCl and then the calcium is precipitated as CaC_2O_4 After filtering and washing the precipitate, it requires 40.0 mL of 0.250 N KMnO_4solution acidified with H_2SO_4to titrate it as, MnO_(4)^(-)+H^(+) +C_(2)O_(4)^(2-) rarr Mn^(2+)+CO_(2)+2H_(2)O The percentage of CaO in the sample is:
Answer» <html><body><p>` 54.0% `<br/>`27.1% `<br/>` <a href="https://interviewquestions.tuteehub.com/tag/42-316129" style="font-weight:bold;" target="_blank" title="Click to know more about 42">42</a>%`<br/>` 84% `</p>Solution :No of <a href="https://interviewquestions.tuteehub.com/tag/milli-560824" style="font-weight:bold;" target="_blank" title="Click to know more about MILLI">MILLI</a> <a href="https://interviewquestions.tuteehub.com/tag/equivalents-974752" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENTS">EQUIVALENTS</a> of lime <a href="https://interviewquestions.tuteehub.com/tag/stone-1228007" style="font-weight:bold;" target="_blank" title="Click to know more about STONE">STONE</a> <br/> = no of milli equivalents of `CaC_(2)O_4` <br/> = noof milli equivalents of`KMnO_4` <br/> `= 40xx0.25 =10` <br/> No of milli moles of lime = 5 <br/>`:.` weight of `CaO = 5 xx 56 xx 10^(-3) = 0.28g` <br/>percentage of `CaO= (0.28)/(0.518)xx100=54.05%`</body></html>