`12.825 gm` of a sample of `Ba(OH)_(2)` is dissolved in `10 ml` of `0.5 N HCl` solution. The excess of `HCl` was titrated with `0.2 N NaOH`. The volume of `NaOH` used was `10 c c`. The percentage of `Ba(OH)_(2)` in the sample is

Options

A. (a)`2.58`
B. (b)`6.4`
C. (c )`8`
D. (d)`2`

Answer

D

Explanation

`HCl" " NaOH`
`V_(1)N_(1)=V_(2)N_(2)`
`V_(1)xx0.5=10xx0.2`
`V_(1)=4 ml`
Vol. of HCl reacted with `Ba(OH)_(2)=10-4=6 ml`
`HCl" " Ba(OH)_(2)`
`VN=W/E`
`6/1000xx0.5=W/85.5implies W=51.3/200`
`% of Ba(OH)_(2)=51.3/(200xx12.825)xx100= 2%`