Equal volumes of solutions in two different vessels are represented above [VISUAL]. If the solution — Solutions Chemistry Question

STEPS:

1. Analyze the particulate representation of Vessel 1 (KCl):
* Vessel 1 represents an aqueous solution of $\text{KCl}$.
* $\text{KCl}$ is a strong electrolyte that dissociates completely into its constituent ions in water:
$$\text{KCl}(aq) \rightarrow \text{K}^+(aq) + \text{Cl}^-(aq)$$
* Counting the ions shown in Vessel 1:
* Positive ions ($\text{K}^+$, represented by black circles): 5 particles.
* Negative ions ($\text{Cl}^-$, represented by white circles): 5 particles.
* This represents a $1:1$ ratio of cations to anions, maintaining electrical neutrality. The molarity of this solution is proportional to the 5 formula units of $\text{KCl}$ dissolved in this volume.

2. Analyze the particulate representation of Vessel 2:
* Counting the ions shown in Vessel 2:
* Positive ions (black circles): 10 particles.
* Negative ions (white circles): 20 particles.
* This represents a $1:2$ ratio of positive ions to negative ions (10 positive to 20 negative).

3. Identify the chemical formula of the solute in Vessel 2:
* Because the solution must be electrically neutral, the total positive charge must equal the total negative charge.
* A $1:2$ ratio of cations to anions indicates that each cation must carry a $+2$ charge to balance two anions each carrying a $-1$ charge (representing an $\text{XY}_2$ salt).
* Between the given options, $\text{CaCl}_2$ is a soluble ionic salt that dissociates in a $1:2$ ratio:
$$\text{CaCl}_2(aq) \rightarrow \text{Ca}^{2+}(aq) + 2\ \text{Cl}^-(aq)$$
* Therefore, the solute in Vessel 2 must be $\text{CaCl}_2$.

4. Determine the relative molarity of Vessel 2 compared to Vessel 1:
* Both vessels contain equal volumes of solution.
* The molarity ($M$) of a solute is directly proportional to the number of formula units (or moles of the cation) dissolved in that volume:
* For $\text{KCl}$ in Vessel 1: $\text{Molarity} \propto 5\text{ formula units}$ (or 5 $\text{K}^+$ ions).
* For $\text{CaCl}_2$ in Vessel 2: $\text{Molarity} \propto 10\text{ formula units}$ (or 10 $\text{Ca}^{2+}$ ions).
* Comparing the concentration of the salts:
$$\frac{\text{Molarity of Vessel 2}}{\text{Molarity of Vessel 1}} = \frac{10}{5} = 2$$
* This indicates that Vessel 2 contains a $\text{CaCl}_2$ solution with twice the molarity of the $\text{KCl}$ solution in Vessel 1, identifying Option D as the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect: If Vessel 2 were a $\text{KCl}$ solution with the same molarity as Vessel 1, it would contain the exact same number of particles (5 black circles and 5 white circles).
• Option B is incorrect: If Vessel 2 were a $\text{KCl}$ solution with twice the molarity, it would contain twice as many formula units of $\text{KCl}$ (10 $\text{K}^+$ and 10 $\text{Cl}^-$), which would be represented as 10 black circles and 10 white circles. However, the diagram shows 20 white circles.
• Option C is incorrect: If Vessel 2 were a $\text{CaCl}_2$ solution with the same molarity as Vessel 1, it would contain the same number of formula units (5 units), which would yield 5 $\text{Ca}^{2+}$ ions (5 black circles) and 10 $\text{Cl}^-$ ions (10 white circles). This is exactly half the number of particles illustrated in Vessel 2.