What is the molarity of I -(aq) in a solution that contains 34 g of SrI2 (molar mass 341 g) in 1.0 L — Solutions Chemistry Question

STEPS:

1. Identify the mass and molar mass of the solute: The problem states that the solution contains $34\text{ g}$ of strontium iodide ($\text{SrI}_2$), which has a molar mass of $341\text{ g/mol}$.
2. Calculate the moles of solute: Use the mass-to-mole conversion formula to find the moles of $\text{SrI}_2$ in the sample:
$$\text{Moles of }\text{SrI}_2 = \frac{34\text{ g}}{341\text{ g/mol}} \approx \mathbf{0.10\text{ mol}}$$
3. Determine the molarity of the $\text{SrI}_2$ solution: Molarity ($M$) is defined as the moles of solute per liter of solution. Since the total volume of the solution is $1.0\text{ L}$:
$$\text{Molarity of }\text{SrI}_2 = \frac{0.10\text{ mol}}{1.0\text{ L}} = \mathbf{0.10\text{ M}}$$
4. Analyze the dissociation of strontium iodide: Strontium iodide is a soluble ionic salt and a strong electrolyte. In an aqueous solution, it dissociates completely into its constituent ions:
$$\text{SrI}_2(aq) \rightarrow \text{Sr}^{2+}(aq) + 2\text{ I}^-(aq)$$
This stoichiometric relationship shows that for every 1 mole of dissolved $\text{SrI}_2$, there are 2 moles of aqueous iodide ($\text{I}^-$) ions released into the solution.
5. Calculate the final molarity of $\text{I}^-$ ions: Multiply the molarity of the parent salt solution by the stoichiometric factor of 2:
$$\text{Molarity of }\text{I}^- = 2 \times 0.10\text{ M} = \mathbf{0.20\text{ M}}$$
This confirms that the correct answer is Option D.

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WHY_OTHERS_WRONG:

• A is incorrect: $0.034\text{ M}$ represents a severe calculation or decimal placement error. A student might arrive at this value by dividing the mass of the salt ($34\text{ g}$) directly by an incorrect factor rather than converting to moles first.
• B is incorrect: $0.068\text{ M}$ is double the incorrect calculation in Option A and has no valid chemical basis under these solution parameters.
• C is incorrect: $0.10\text{ M}$ represents the molarity of the $\text{SrI}_2$ salt itself, or the molarity of strontium ions ($\text{Sr}^{2+}$). A student selecting this option forgot to multiply by the stoichiometric coefficient of 2 for the iodide ions released during complete dissociation.
• E is incorrect: $0.68\text{ M}$ represents a substantial arithmetic error, possibly from incorrectly scaling the molar mass of the salt or misplaced decimals during long division.