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$1.387g$ of a sample containing KCl and $NH_4Cl$ is heated until constant weight.The residue is dissolved in 20ml of N/10 $AgNO_3$ solution.Calculate % of chlorine in mixture.

$(a)\;71.2\%\qquad(b)\;73.2\%\qquad(c)\;61.2\%\qquad(d)\;64.31\%$

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Weight of KCl='a'g
Weight of KCl='b'g
$a+b=1.387$-------(1)
On heating
$KCl\rightarrow $No change
$NH_4Cl\rightarrow NH_3\uparrow +HCl \uparrow$
The residue is only KCl,which reacts with $AgNO_3$
Meq of KCl=Meq of $AgNO_3$
$\large\frac{a}{74.5}$$\times 1000$$=20\times \large\frac{1}{10}$
$a=w_{KCl}=0.149g$
$b=w_{NH_4Cl}=1.387-0.149g=1.238g$
$\large\frac{w}{35.5}$$\times 1000=\large\frac{0.149}{74.5}$$\times 1000+\large\frac{1.238}{53.5}$$\times 1000$
$w_{Cl}=0.892g$
% of $Cl$ in mixture=$\large\frac{0.892}{1.387}$$\times 100$
$\Rightarrow 64.31\%$
Hence (d) is the correct answer.
answered Jan 7, 2014 by sreemathi.v
 

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