Q.56. The solubility product constants of AgCrO and AgBr are 32x and 4y respectively at 298 K. The v

Q: What is the correct answer for this exam Chemistry Q56?

The correct answer is D. For Ag$_2$CrO$_4$: $\mathrm{Ag_2CrO_4(s)} \rightleftharpoons 2\mathrm{Ag^+(aq)} + \mathrm{CrO_4^{2-}(aq)}$. If $S_1$ is the molar solubility of Ag$_2$CrO$_4$, then $[\mathrm{Ag^+}] = 2S_1$ and $[\mathrm{CrO_4^{2-}}] = S_1$. $K_{sp}(\mathrm{Ag_2CrO_4}) = (2S_1)^2 (S_1) = 4S_1^3$. Given $K_{sp}(\mathrm{Ag_2CrO_4}) = 32x$. So, $4S_1^3 = 32x \Rightarrow S_1^3 = 8x \Rightarrow S_1 = (8x)^{1/3} = 2x^{1/

Q: Q.56. The solubility product constants of AgCrO and AgBr are 32x and 4y respectively at 298 K. The value of can be expressed as :

The correct answer is D. For Ag$_2$CrO$_4$: $\mathrm{Ag_2CrO_4(s)} \rightleftharpoons 2\mathrm{Ag^+(aq)} + \mathrm{CrO_4^{2-}(aq)}$. If $S_1$ is the molar solubility of Ag$_2$CrO$_4$, then $[\mathrm{Ag^+}] = 2S_1$ and $[\mathrm{CrO_4^{2-}}] = S_1$. $K_{sp}(\mathrm{Ag_2CrO_4}) = (2S_1)^2 (S_1) = 4S_1^3$. Given $K_{sp}(\mathrm{Ag_2CrO_4}) = 32x$. So, $4S_1^3 = 32x \Rightarrow S_1^3 = 8x \Rightarrow S_1 = (8x)^{1/3} = 2x^{1/

Q: What are the options for this this exam Chemistry question?

A. $2\sqrt[3]{x}/y$ · B. $2\sqrt{y}/x$ · C. $x/y$ · D. $\sqrt[3]{x}/\sqrt{y}$

Question

Q.56. The solubility product constants of AgCrO and AgBr are 32x and 4y respectively at 298 K. The value of  can be expressed as :

FirstInTest · Accepted Answer

For Ag$_2$CrO$_4$: $\mathrm{Ag_2CrO_4(s)} ightleftharpoons 2\mathrm{Ag^+(aq)} + \mathrm{CrO_4^{2-}(aq)}$. If $S_1$ is the molar solubility of Ag$_2$CrO$_4$, then $[\mathrm{Ag^+}] = 2S_1$ and $[\mathrm{CrO_4^{2-}}] = S_1$. $K_{sp}(\mathrm{Ag_2CrO_4}) = (2S_1)^2 (S_1) = 4S_1^3$. Given $K_{sp}(\mathrm{Ag_2CrO_4}) = 32x$. So, $4S_1^3 = 32x \Rightarrow S_1^3 = 8x \Rightarrow S_1 = (8x)^{1/3} = 2x^{1/3}$. The molarity of Ag$_2$CrO$_4$ in saturated solution is $S_1$.
For AgBr: $\mathrm{AgBr(s)} ightleftharpoons \mathrm{Ag^+(aq)} + \mathrm{Br^-(aq)}$. If $S_2$ is the molar solubility of AgBr, then $[\mathrm{Ag^+}] = S_2$ and $[\mathrm{Br^-}] = S_2$. $K_{sp}(\mathrm{AgBr}) = S_2^2$. Given $K_{sp}(\mathrm{AgBr}) = 4y$. So, $S_2^2 = 4y \Rightarrow S_2 = (4y)^{1/2} = 2y^{1/2}$. The molarity of AgBr in saturated solution is $S_2$.
We need to find the ratio $\frac{	ext{molarity of Ag}_2	ext{CrO}_4}{	ext{molarity of AgBr}} = \frac{S_1}{S_2}$.
$\frac{S_1}{S_2} = \frac{2x^{1/3}}{2y^{1/2}} = \frac{x^{1/3}}{y^{1/2}} = \frac{\sqrt[3]{x}}{\sqrt{y}}$. This matches option (4).

Answer

$2\sqrt[3]{x}/y$

Answer

$2\sqrt{y}/x$

Answer

$x/y$

Answer

$\sqrt[3]{x}/\sqrt{y}$

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