conductivity lab check (revised)\nwhich of the following ionic compounds will produce the most ions when it…

conductivity lab check (revised)\nwhich of the following ionic compounds will produce the most ions when it dissociates in water?\n ○ cacl₂\n ◉ mg(no₃)₂\n ○ kbr\n ○ alf₃

conductivity lab check (revised)\nwhich of the following ionic compounds will produce the most ions when it dissociates in water?\n ○ cacl₂\n ◉ mg(no₃)₂\n ○ kbr\n ○ alf₃

Answer

Explanation:

Step1: Analyze each compound's dissociation

For $\ce{CaCl2}$: Dissociates as $\ce{CaCl2 -> Ca^{2+} + 2Cl^-}$, total ions per formula unit: $1 + 2 = 3$. For $\ce{Mg(NO3)2}$: Dissociates as $\ce{Mg(NO3)2 -> Mg^{2+} + 2NO3^-}$, total ions per formula unit: $1 + 2 = 3$. Wait, no, wait—wait, $\ce{AlF3}$: $\ce{AlF3 -> Al^{3+} + 3F^-}$, total ions: $1 + 3 = 4$. Wait, maybe the initial selection was wrong. Wait, let's recalculate:

$\ce{CaCl2}$: 1 $\ce{Ca^{2+}}$ and 2 $\ce{Cl^-}$, total 3 ions.

$\ce{Mg(NO3)2}$: 1 $\ce{Mg^{2+}}$ and 2 $\ce{NO3^-}$, total 3 ions.

$\ce{KBr}$: 1 $\ce{K^+}$ and 1 $\ce{Br^-}$, total 2 ions.

$\ce{AlF3}$: 1 $\ce{Al^{3+}}$ and 3 $\ce{F^-}$, total 4 ions. Wait, so the correct answer should be $\ce{AlF3}$? Wait, maybe the original problem's selected option was incorrect. Wait, let's check again.

Wait, the question is "which will produce the most ions when it dissociates in water". Let's calculate the number of ions per formula unit:

  • $\ce{CaCl2}$: $\ce{CaCl2 (s) -> Ca^{2+}(aq) + 2Cl^-(aq)}$ → 3 ions.
  • $\ce{Mg(NO3)2}$: $\ce{Mg(NO3)2 (s) -> Mg^{2+}(aq) + 2NO3^-(aq)}$ → 3 ions.
  • $\ce{KBr}$: $\ce{KBr (s) -> K^+(aq) + Br^-(aq)}$ → 2 ions.
  • $\ce{AlF3}$: $\ce{AlF3 (s) -> Al^{3+}(aq) + 3F^-(aq)}$ → 4 ions.

So $\ce{AlF3}$ produces 4 ions, which is more than 3 (from $\ce{CaCl2}$ and $\ce{Mg(NO3)2}$) and 2 (from $\ce{KBr}$). So the correct answer should be $\ce{AlF3}$. But the original selected option was $\ce{Mg(NO3)2}$, which is wrong. Wait, maybe I made a mistake? Wait, no—wait, is $\ce{AlF3}$ soluble? Well, the question is about dissociation, assuming they dissociate completely. So based on the number of ions per formula unit, $\ce{AlF3}$ gives 4 ions, which is the most. So the correct option is $\ce{AlF3}$.

Wait, but let's confirm:

$\ce{CaCl2}$: 3 ions.

$\ce{Mg(NO3)2}$: 3 ions.

$\ce{KBr}$: 2 ions.

$\ce{AlF3}$: 4 ions. So yes, $\ce{AlF3}$ produces the most ions. So the correct answer is $\ce{AlF3}$, so the option is D (assuming $\ce{AlF3}$ is option D). Wait, the options are:

  • $\ce{CaCl2}$ (option 1)
  • $\ce{Mg(NO3)2}$ (option 2)
  • $\ce{KBr}$ (option 3)
  • $\ce{AlF3}$ (option 4)

So the correct answer is $\ce{AlF3}$, so the option is the last one, $\ce{AlF3}$.

Answer:

$\ce{AlF3}$ (the option with $\ce{AlF3}$)