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» `1.5%` `2.6%` `3.4%` `1.3%` Solution :MILLI eq. of HCl initially = `10xx0.5=5` Milli eq of NaOH consumed = milli eq. of HCl in EXCESS = `10xx0.2-2` `THEREFORE` m.eq of HCl consumed = milli eq of `Ba(OH)_(2)=5-2=3` Thus, equire of `Ba(OH)_(2)=3XX10^(-3)` Mass of `Ba(OH)_(2)` = Equivalents `xx` Eq. wt `=3xx10^(-3)xx(171//2)` `=0.2565gm` `therefore % Ba(OH)_(2)=(0.2565)/(20)xx100=1.28%`