What
is
the
molality
of
lithium
ions
in
a
0.302
m
solution
of
Li₃PO₄
assuming
the
compound
dissociates
completely?

To calculate the molality of lithium ions (Li⁺) in a 0.302 m (molal) solution of lithium phosphate (Li₃PO₄), follow these steps:

1. **Dissociation of Li₃PO₄**: Lithium phosphate dissociates completely in water into lithium ions (Li⁺) and phosphate ions (PO₄³⁻) as per the equation:
   \[
   \text{Li₃PO₄ (s)} \rightarrow 3 \text{Li⁺ (aq)} + \text{PO₄³⁻ (aq)}
   \]
   For every 1 mole of Li₃PO₄, it produces 3 moles of lithium ions (Li⁺).

2. **Molality of lithium ions (Li⁺)**: Since 1 mole of Li₃PO₄ gives 3 moles of Li⁺, the molality of lithium ions in the solution is:
   \[
   \text{Molality of Li⁺} = 3 \times \text{molality of Li₃PO₄}
   \]
   Given the molality of Li₃PO₄ is 0.302 m, the molality of lithium ions is:
   \[
   \text{Molality of Li⁺} = 3 \times 0.302 = 0.906 \, \text{m}
   \]

Thus, the molality of lithium ions (Li⁺) is **0.906 m**.

Would you like further clarification or details on this calculation?

Here are 5 related questions for further exploration:
1. What is the molality of phosphate ions in the solution?
2. How does the concept of molality differ from molarity?
3. What is the van't Hoff factor for complete dissociation of Li₃PO₄?
4. How would you calculate the freezing point depression for this solution?
5. What would happen to the molality if the solution were diluted?

**Tip:** Molality is temperature-independent, making it particularly useful in experiments involving temperature changes!

What is the molality of lithium ions in a 0.302 m solution of Li₃PO₄ assuming the compound dissociates completely?

Discover how to calculate the molality of lithium ions in a 0.302 m solution of lithium phosphate (Li₃PO₄). This chemistry problem involves dissociation, stoichiometry, and molality calculations, making it suitable for Grades 10-12.

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