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A solution contains 4g of $Na_2CO_3$ and NaCl 250ml.25ml of this solution required 50ml of N/10 HCl for complete neutralization.Calculate % composition of mixture.

$\begin{array}{1 1}(a)\;66.25\%,33.75\%&(b)\;60\%,40\%\\(c)\;60.25\%,40.75\%&(d)\;65\%,35\%\end{array}$

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Weight of $Na_2CO_3$='a'g
Weight of $NaCl$='b'g
$a+b=4$----(1)
The solution reacts with HCl,only $Na_2CO_3$ reacts.
$\therefore$ In 25ml,Meq of $Na_2CO_3$=Meq of HCl
$\Rightarrow 50\times \large\frac{1}{10}$
$\Rightarrow 5$
$\therefore$Meq of $Na_2CO_3$ in 250ml=$\large\frac{5\times 250}{25}$
$\Rightarrow 50$
$\large\frac{a}{\Large\frac{106}{2}}$$\times 1000=50$
$a=2.65$
From equ(1)
$b=4-2.65=1.35g$
% of $Na_2CO_3=\large\frac{2.65}{4}$$\times 100=66.25\%$
$\therefore$ % of NaCl=33.75%
Hence (a) is the correct answer.
answered Jan 7, 2014 by sreemathi.v
 

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