Question

$6$ g mixture of $N{H}_{4}Cl$ and $NaCl$ is treated with $110$ mL of a solution of caustic soda of $0.63$ N. The solution is then boiled to remove $N{H}_3$. The resulting solution requires $48.1$ mL of a solution of the $0.1$ N $HCl$. The $\%$ composition of the mixture is (only integer part) :

Answer 1

$NaOH$ added in mixture of $N{H}_{4}Cl$ and $NaCl$ reacts with $N{H}_{4}Cl$ only to give off $N{H}_3$
The excess of $NaOH$ is used by $HCl$.
Meq. of $NaOH$ added in mixture $=110\times 0.63=69.3$
Meq. of $NaOH$ left after reaction with $N{H}_{4}Cl=$Meq. of $HCl$ used $=48.1\times 0.1=4.81$
Now, Meq. of $N{H}_{4}Cl=$Meq. of $NaOH$ used for $N{H}_{4}Cl$ $=$$69.3-4.81=$$64.49$
$\therefore \left(\frac{w}{53.5}\right)\times 1000=64.49$
or, $w_{N{H}_{4}Cl}=\left(\frac{3.45}{6}\right)\times 100=57.5$%
and $\%$ of $NaCl$ $=$$100-57.5$$=42.5$ %
So, answer is $42$.