Suppose 23.8 g of sodium iodide is dissolved in 350. mL of a 0.80 M aqueous solution of silver nitrate. Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Round your answer to 3 significant digits. M D ? x

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### Problem Statement

Suppose 23.8 g of sodium iodide is dissolved in 350. mL of a 0.80 M aqueous solution of silver nitrate.

Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it.

**Round your answer to 3 significant digits.**

### Solution Explanation

To solve this problem, follow these steps:

1. **Determine moles of sodium iodide (NaI):**
   - Use the molar mass of sodium iodide to convert the given mass to moles.
   
     \( \text{Molar mass of NaI} = 22.99 \, (\text{Na}) + 126.90 \, (\text{I}) = 149.89 \, \text{g/mol} \)
     
     \( \text{Moles of NaI} = \frac{23.8 \, \text{g}}{149.89 \, \text{g/mol}} \approx 0.159 \, \text{mol} \)

2. **Determine the volume of the solution in liters:**
   - Convert the volume of the solution from milliliters to liters.
   
     \( 350. \, \text{mL} = 0.350 \, \text{L} \)

3. **Calculate the molarity of sodium cation in the solution:**
   - Sodium iodide dissociates completely in water into \( \text{Na}^+ \) and \( \text{I}^- \).
   - Therefore, the moles of sodium ions (\( \text{Na}^+ \)) will be equal to the moles of sodium iodide.
   - Molarity (\( \text{M} \)) is defined as moles of solute per liter of solution.
   
     \( \text{Molarity of Na}^+ = \frac{\text{Moles of NaI}}{\text{Volume of solution in L}} = \frac{0.159 \, \text{mol}}{0.350 \, \text{L}} \approx 0.454 \, \text{M} \)
 
### Input Field

\[ \boxed{ \text{M}} \]

### Additional Features:
- A checkbox labeled "x10"
Transcribed Image Text:### Problem Statement Suppose 23.8 g of sodium iodide is dissolved in 350. mL of a 0.80 M aqueous solution of silver nitrate. Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. **Round your answer to 3 significant digits.** ### Solution Explanation To solve this problem, follow these steps: 1. **Determine moles of sodium iodide (NaI):** - Use the molar mass of sodium iodide to convert the given mass to moles. \( \text{Molar mass of NaI} = 22.99 \, (\text{Na}) + 126.90 \, (\text{I}) = 149.89 \, \text{g/mol} \) \( \text{Moles of NaI} = \frac{23.8 \, \text{g}}{149.89 \, \text{g/mol}} \approx 0.159 \, \text{mol} \) 2. **Determine the volume of the solution in liters:** - Convert the volume of the solution from milliliters to liters. \( 350. \, \text{mL} = 0.350 \, \text{L} \) 3. **Calculate the molarity of sodium cation in the solution:** - Sodium iodide dissociates completely in water into \( \text{Na}^+ \) and \( \text{I}^- \). - Therefore, the moles of sodium ions (\( \text{Na}^+ \)) will be equal to the moles of sodium iodide. - Molarity (\( \text{M} \)) is defined as moles of solute per liter of solution. \( \text{Molarity of Na}^+ = \frac{\text{Moles of NaI}}{\text{Volume of solution in L}} = \frac{0.159 \, \text{mol}}{0.350 \, \text{L}} \approx 0.454 \, \text{M} \) ### Input Field \[ \boxed{ \text{M}} \] ### Additional Features: - A checkbox labeled "x10"
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