1-387 g of a sample containing KCl and NH4Cl is heated until constant mass- The residue is dissolved in 20 mL of N-10 AgNO3 solution- The - of chlorine in the mixture is as the nearest integer -

Let the mass of KCl and NH_4Cl in a mixture nbsp;be a and b respectively. ∴ nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Standard XII

Chemistry

Stoichiometric Calculations

1.387 g of a ...

Question

$1.387$ g of a sample containing $K C l$ and $N H_{4} C l$ is heated until constant mass. The residue is dissolved in $20$ mL of $\frac{N}{10} A g N O_{3}$ solution. The $%$ of chlorine in the mixture is (as the nearest integer) :

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Solution

Let the mass of $K C l$ and $N H_{4} C l$ in a mixture be $a$ and $b$ respectively.
$∴ a + b = 1.387 . . . (i)$
$N H_{4} C l Δ - \to N H_{3} ↑ + H C l ↑$
On heating $K C l$ remains unaffected, whereas $N H_{3} C l$ is decomposed to give off $N H_{3}$ and $H C l$ .
The residue left is $K C l$ only which reacts with $A g N O_{3}$ .
Meq. of $K C l$ = Meq. of $A g N O_{3}$
$\frac{a}{74.5} \times 1000 = 20 \frac{1}{10}$
$∴ a = 0.149 g$ (mass of $K C l$ )
By Eq. (i), $∴ b = 1.387 - 0.149 = 1.238 g$ (mass of $N H_{4} C l$
Now for mass of ${C l}_{2}$ in mixture
Meq. of ${C l}_{2} =$ Meq. of KCl+ Meq. of $N H_{4} C l$
$\frac{W_{c l}}{35.5} \times 1000 = \frac{0.149}{74.5} \times 1000 + \frac{1.238}{53.5} \times 1000$
$W_{C l} = 0.892 g$
$%$ of chlorine in mixture $= \frac{0.892}{1.387} \times 100 = 64.35 %$ .
So, the answer is $64$ .

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