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

Options

A. 4
B. 12
C. 8
D. 16

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 for molality is:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]

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

### Step 3: Calculate the number of moles of Na⁺ ions
To calculate the number of moles of Na⁺, we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
The molar mass of sodium (Na) is approximately 23 g/mol. Therefore:
\[ \text{Number of moles of Na⁺} = \frac{92 \text{ g}}{23 \text{ g/mol}} \]
Calculating this gives:
\[ \text{Number of moles of Na⁺} = 4 \text{ moles} \]

### Step 4: Calculate the molality
Now that we have the number of moles, we can substitute this value into the molality formula:
\[ m = \frac{4 \text{ moles}}{1 \text{ kg}} \]
Thus, the molality of Na⁺ ions in the solution is:
\[ m = 4 \text{ mol/kg} \]

### Final Answer
The molality of Na⁺ ions in the solution is **4 mol/kg**.

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