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Question

Solubility of $$AgCl$$ in $$0.2\ M\ NaCl$$ is $$x$$ and that in $$0.1\ M\ AgNO_{3}$$ is $$y$$ then which of the following is correct?


A
x=y
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B
x>y
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C
y>x
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D
Cannot be predicted
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Solution

The correct option is B $$y>x$$
In $$0.2M$$ $$NaCl$$, $$[Cl^-]=0.2$$
$$(K_{sp})_{AgCl}=[Ag^+][Cl^-]$$
$$\therefore (K_{sp})_{AgCl}=x\times 0.2$$
$$\Rightarrow x=\cfrac {(K_{sp})_{AgCl}}{0.2}=5(K_{sp})_{AgCl}$$
In $$0.1M$$ $$AgNO_3$$, $$[Ag^+]=0.1M$$
$$\Rightarrow (K_{sp})_{AgCl}=0.1\times y$$
$$\Rightarrow y=\cfrac {(K_{sp})_{AgCl}}{0.1}=10(K_{sp})_{AgCl}$$
$$\therefore y >x$$

Chemistry

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