$20g$  of sample of $Ba{\left(OH\right)}_2$  is dissolved in $10mL$ . Of  0.5  $NHCl$ solution. The excess of $HCl$  was titrated with $0.2NNaOH$ . The volume of $NaOH$ used was 10 cc. Calculate the percentage of  $Ba{\left(OH\right)}_2$ in the sample.

Milli eq. of $HCl$  initially  $=10\times 0.5=5$
Milli eq. of $NaOH$  consumed $=$  Milli eq. of $HCl$  in excess
                                            $=10\times 0.2=2$
$\therefore$ Milli eq  $HCl$ . consumed = Milli eq. of  $Ba{\left(OH\right)}_2$
                                           $=5-2=3$
 $\therefore eq.ofBa{\left(OH\right)}_2=\frac{3}{1000}=3\times {10}^{-3}$
Mass of  $Ba{\left(OH\right)}_2=3\times {10}^{-3}\times (171/2)=0.02565g$
 $\%Ba{\left(OH\right)}_2=(0.2565/20)\times 100=1.28\%$