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

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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.

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