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20g of sample of $Ba(OH)_2$ is dissolved in 50c.c of 0.1N HCl solution.The excess of HCl has titrated with 0.1NaOH.The volume of NaOH used was 20c.c.Calculate percentage of $Ba(OH)_2$ in the sample

$\begin{array}{1 1}(a)\;1.28\%& (b)\;12.8\% \\(c)\;25.6\%&(d)\;0.128\%\end{array}$

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Eq.of $Ba(OH)_2$(pure)=Eq.of HCl used
$\Rightarrow $Total Eq.of HCl taken-eq.of HCl left
$\Rightarrow$ Eq.of HCl taken-eq.of NaOH used for back titration.
$\large\frac{W}{171/2}$$=50\times 0.1\times 10^{-3}-0.1\times 20\times 10^{-3}$
$W=0.256$gm
Percentage purity of $Ba(OH)_2$ in the sample =$\large\frac{0.256}{20}$$\times 100=1.28\%$
$\therefore$ The option is (a) 1.28%
answered Mar 24, 2014 by sreemathi.v
 

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