What is the molality of lithium ions in a 0.302 m solution of Li\(_3\)PO\(_4\), assuming the compound dissociates completely?

• A. 1.208 m
• B. 0.302 m
• C. 0.906 m
• D. 0.604 m

Answer: (C)

Verified step by step guidance

1. Understand that molality (m) is defined as moles of solute per kilogram of solvent. Here, the given molality (0.302 m) refers to the molality of Li₃PO₄, the solute.
2. Recognize that Li₃PO₄ dissociates completely in water into lithium ions (Li⁺) and phosphate ions (PO₄³⁻). The dissociation equation is: \(\mathrm{Li_3PO_4 \rightarrow 3Li^+ + PO_4^{3-}}\).
3. Since each formula unit of Li₃PO₄ produces 3 lithium ions, the molality of lithium ions will be 3 times the molality of Li₃PO₄.
4. Calculate the molality of lithium ions by multiplying the given molality of Li₃PO₄ by 3: \(m_{Li^+} = 3 \times 0.302\,m\).
5. Interpret the result as the molality of lithium ions in the solution, which reflects the total moles of Li⁺ per kilogram of solvent after complete dissociation.