What is the ionic molality of sodium ions in a solution of 25.7 g NaNO₃ dissolved in enough water to make a 150.0 mL of solution? Density of the solution is 1.02 g/mL.

• A. 1.98 m
• B. 2.57 m
• C. 2.98 m
• D. 2.37 m

Answer: (D)

Verified step by step guidance

1. Calculate the number of moles of sodium nitrate (NaNO\_3) using its molar mass. Use the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). The molar mass of NaNO\_3 is the sum of the atomic masses of Na, N, and 3 O atoms.
2. Determine the total mass of the solution by multiplying the volume of the solution by its density: \(\text{mass of solution} = \text{volume} \times \text{density}\). Make sure to convert the volume from mL to grams using the density given.
3. Calculate the mass of the solvent (water) by subtracting the mass of NaNO\_3 from the total mass of the solution: \(\text{mass of solvent} = \text{mass of solution} - \text{mass of solute}\).
4. Convert the mass of the solvent from grams to kilograms because molality is expressed in moles of solute per kilogram of solvent: \(\text{mass of solvent (kg)} = \frac{\text{mass of solvent (g)}}{1000}\).
5. Calculate the molality of sodium ions. Since each formula unit of NaNO\_3 dissociates into one Na\^+ ion, the moles of sodium ions equal the moles of NaNO\_3. Use the formula for molality: \(m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}\). This gives the ionic molality of sodium ions.