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20 gm of sample Ba(OH)(2) is dissolved i...

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.

A

`1.5%`

B

`2.6%`

C

`3.4%`

D

`1.3%`

Text Solution

Verified by Experts

The correct Answer is:
D
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%`
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