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

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%

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