20 gm of sample Ba(OH)_(2) is dissolved in 10 ml of 0.5 N HCl solution, The excess of HCl was titrated with 0.2 NaOH. The volume of NaOh used was 10 mol. Calculate the % of Ba(OH)_(2) in the sample.

20 gm of sample Ba(OH)_(2) is dissolved in 10 ml of 0.5 N HCl solution

1.	20 gm of sample Ba(OH)_(2) is dissolved in 10 ml of 0.5 N HCl solution, The excess of HCl was titrated with 0.2 NaOH. The volume of NaOh used was 10 mol. Calculate the % of Ba(OH)_(2) in the sample.
Answer» <html><body><p>`1.5%`<br/>`2.6%`<br/>`3.4%`<br/>`1.3%`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/milli-560824" style="font-weight:bold;" target="_blank" title="Click to know more about MILLI">MILLI</a> eq. of HCl initially = `10xx0.5=5` <br/> Milli eq of NaOH consumed <br/> = milli eq. of HCl in <a href="https://interviewquestions.tuteehub.com/tag/excess-978535" style="font-weight:bold;" target="_blank" title="Click to know more about EXCESS">EXCESS</a> = `10xx0.2-2` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` m.eq of HCl consumed <br/> = milli eq of `Ba(OH)_(2)=5-2=3` <br/> Thus, equire of `Ba(OH)_(2)=<a href="https://interviewquestions.tuteehub.com/tag/3xx10-1865443" style="font-weight:bold;" target="_blank" title="Click to know more about 3XX10">3XX10</a>^(-3)` <br/> Mass of `Ba(OH)_(2)` = Equivalents `xx` Eq. wt <br/> `=3xx10^(-3)xx(171//2)` <br/> `=0.2565gm` <br/> `therefore % Ba(OH)_(2)=(0.2565)/(20)xx100=1.28%`</body></html>