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If $K_{sp}$ of $Ag_{2}CO_{3}$ is $8 \times 10^{-12}$ , the molar solubility of $Ag_{2}CO_{3}$ in 0.1 M $AgNO_{3}$ is :
Option 1)
$8 \times 10^{-11}$ M

Option 2)
$8 \times 10^{-12}$ M
Option 3)
$8 \times 10^{-10}$ M
Option 4)
$8 \times 10^{-13}$ M

Answers (1)

best_answer

Relation between concentration terms (Molality & Mole fraction) -

$m= \frac{x_{solute}\times 1000}{x_{solvent}\times M_{1}}$

- wherein

$M_{1}= Molecular\ Mass\ of \ Solvent$

$x_{solute}= mole\: f\! raction\ of\ solute$

$x_{solvent}= mole\: f\! raction\ of\ solvent$

Relation between concentration terms (Molality & Molarity) -

$\dpi{100} m=\frac{1000M}{1000d-Mm_{1}}$

- wherein

$m=Molality$

$M=Molarity$

$m_{1}=Molar\: mass\: of \: solute$

$\dpi{100} d=density\: of \: solution \: g/ml$

As we have learned in solution

The reaction is

$Ag_{2}CO_{3}(s)\rightleftharpoons 2Ag+(aq)(0.1+25)M+CO{_{3}}^{-2}(aq)(s\: M)$

$Ksp =\left [ Ag^{+} \right ]^{2}\left [ CO{_{3}}^{-2} \right ]$

$8\times 10^{-12}=(0.1+25)^{2}(s)$

$s=8\times 10^{-10}M$

Option 1)

$8 \times 10^{-11}$ M

Option 2)

$8 \times 10^{-12}$ M

Option 3)
$8 \times 10^{-10}$ M
Option 4)

$8 \times 10^{-13}$ M

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