A solution of sodium sulfate contains 92g of `Na^(+)` ions per kilogram of water. The molality of `Na^(+)` ions in the solution in mol `kg^(-1)` is

Explanation

To find the molality of sodium ions (Na⁺) in the solution, we can follow these steps:

### Step 1: Understand the definition of molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula is:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

### Step 2: Identify the mass of Na⁺ ions and the mass of water
From the problem, we know that:
- Mass of Na⁺ ions = 92 g
- Mass of water (solvent) = 1 kg (as given in the problem)

### Step 3: Calculate the number of moles of Na⁺ ions
To find the number of moles of Na⁺ ions, we use the molar mass of sodium (Na), which is approximately 23 g/mol. The formula to calculate the number of moles is:
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]

Substituting the values:
\[ \text{Number of moles of Na⁺} = \frac{92 \text{ g}}{23 \text{ g/mol}} = 4 \text{ moles} \]

### Step 4: Calculate the molality of Na⁺ ions
Now that we have the number of moles of Na⁺ ions, we can calculate the molality using the formula from Step 1:
\[ m = \frac{\text{moles of Na⁺}}{\text{mass of water (kg)}} \]

Substituting the values:
\[ m = \frac{4 \text{ moles}}{1 \text{ kg}} = 4 \text{ mol/kg} \]

### Final Answer
The molality of Na⁺ ions in the solution is:
\[ 4 \text{ mol/kg} \]

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