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Henderson Hassleback Equation

2.4 g of a mi...

Question

$2.4$ g of a mixture of $K C l$ and $N H_{4} C l$ is heated till constant weight is obtained. The solid residue is dissolved in water and the solution is made up to $250$ mL. $25$ mL of this solution required $15$ mL of $0.1$ N $A g N O_{3}$ for precipitation of chloride. The percentage of chloride by weight in the mixture is y%, then y is (in nearest integer) :

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Solution

Ammonium chloride on heating will decompose to ammonia and hydrogen chloride which will be lost in the atmosphere. The solid residue will contain $K C l$ only.
$15$ mL of $0.1$ N $A g N O_{3}$ corresponds to $0.015 \times 0.1 = 0.0015 m o l e s$ of silver ions.
They will react with $0.0015$ moles of chloride ion, which are present in $25$ mL of the solution.
$250$ mL of the solution will contain $0.015$ moles of chloride .
The molar mass of $K C l$ is $74.55$ g/mol.
$0.015$ mole of $K C l$ will correspond to $74.55 \times 0.015 = 1.1183 g K C l$ .
The percentage of $K C l$ in the mixture $= \frac{1.1183}{2.4} \times 100 = 46.6$ %
The nearest integer is $47$ .

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