Step 1: Establish the mass of the solution.
A 1 M solution means 1 mole of $ Na_2SO_4 $ per litre of solution. Given density = $ 1.2\text{ g mL}^{-1} $, the mass of 1 L = 1000 mL of solution is: \[ m_{solution} = 1000 \times 1.2 = 1200\text{ g} \]
Step 2: Find the mass of solvent (water).
Molar mass of $ Na_2SO_4 = 2(23) + 32 + 4(16) = 142\text{ g mol}^{-1} $. Mass of 1 mol $ Na_2SO_4 $ = 142 g. Mass of water = $ 1200 - 142 = 1058\text{ g} = 1.058\text{ kg} $.
Step 3: Calculate moles of Na+ from partial dissociation.
The dissociation is: \[ Na_2SO_4 \to 2Na^+ + SO_4^{2-} \] 50% dissociation means 0.5 mol of $ Na_2SO_4 $ dissociates, producing $ 2 \times 0.5 = 1\text{ mol}\ Na^+ $. The remaining 0.5 mol $ Na_2SO_4 $ stays undissociated (it does not contribute free $ Na^+ $).
Step 4: Calculate the molality of Na+ ions.
Molality $ m = \dfrac{\text{moles of species}}{\text{kg of solvent}} $: \[ m_{Na^+} = \frac{1\text{ mol}}{1.058\text{ kg}} \approx 0.945 \approx 0.95\text{ mol kg}^{-1} \]
Step 5: Why molality not molarity?
Molality uses mass of solvent (independent of temperature), while molarity uses volume of solution. The question specifically asks for molality of $ Na^+ $, which accounts for the actual mass of solvent.
Step 6: Final answer.
\[ \boxed{m_{Na^+} \approx 0.95\text{ mol kg}^{-1}} \]