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$1\;g$ of impure $Na_2CO_3$ is dissolved in water and the solution is made upto $250\;ml$. To $50\;ml$ of this made up solution, $50\;ml$ of $0.1\;N\; HCl$ is added and the mixture after shaking well, required $10\;ml$ of $0.16\;N\; NaOH$ solution for complete neutralization. Calculate $\%$ purity of the sample of $Na_2CO_3$

$(a)\;91.2\%\qquad(b)\;93.4\%\qquad(c)\;90.1\%\qquad(d)\;92.3\%$

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Meq of HCl added to 50ml $Na_2CO_3$ solution =$50\times 0.1=5$
Meq of HCl used for $Na_2CO_3$ in 50ml solution = 5 -1.6 = 3.4
$\therefore$ Meq of HCl used for $Na_2CO_3$ in 250ml solution = $\large\frac{3.4\times 250}{50}$$=17$
$\therefore$ Meq of $Na_2CO_3=17$
$\Rightarrow\large\frac{w}{\Large\frac{106}{2}}$$\times 1000=17$
$\Rightarrow w=0.901g$
$\%$ of $Na_2CO_3$ in the sample = $\large\frac{0.901}{1}$$\times 100=90.1$ %
Hence (c) is the correct answer.
answered Jan 7, 2014 by sreemathi.v
edited Mar 18, 2014 by mosymeow_1
 

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