Solubility of $A g C l$ in $0.2 M N a C l$ is $x$ and that in $0.1 M A g N O_{3}$ is $y$ then which of the following is correct?

x = y

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x > y

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y > x

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Cannot be predicted

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Solution

The correct option is B $y > x$
In $0.2 M$ $N a C l$ , $[C l^{-}] = 0.2$
$(K_{s p})_{A g C l} = [A g^{+}] [C l^{-}]$
$∴ (K_{s p})_{A g C l} = x \times 0.2$
$\Rightarrow x = \frac{(K_{s p})_{A g C l}}{0.2} = 5 (K_{s p})_{A g C l}$
In $0.1 M$ $A g N O_{3}$ , $[A g^{+}] = 0.1 M$
$\Rightarrow (K_{s p})_{A g C l} = 0.1 \times y$
$\Rightarrow y = \frac{(K_{s p})_{A g C l}}{0.1} = 10 (K_{s p})_{A g C l}$
$∴ y > x$

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Solubility of AgCl in 0.2 M NaCl is x and that in 0.1 M AgNO3 is y then which of the following is correct?

The correct option is B y>xIn 0.2M NaCl, [Cl−]=0.2(Ksp)AgCl=[Ag+][Cl−]∴(Ksp)AgCl=x×0.2⇒x=(Ksp)AgCl0.2=5(Ksp)AgClIn 0.1M AgNO3, [Ag+]=0.1M⇒(Ksp)AgCl=0.1×y⇒y=(Ksp)AgCl0.1=10(Ksp)AgCl∴y>x

Solubility of $A g C l$ in $0.2 M N a C l$ is $x$ and that in $0.1 M A g N O_{3}$ is $y$ then which of the following is correct?