Final answer:
The molality of a 2.4 m NaCl solution, assuming the solution density is 1.00 g/ml and volume changes are negligible, is approximately 2.4 moles of NaCl per kilogram of water. This is an approximation because the actual mass of the water is slightly less due to the presence of NaCl.
Step-by-step explanation:
To calculate the molality of a 2.4 m NaCl solution (where m stands for molarity, not molality), we need to convert molarity to molality. Since the density of the solution is given as 1.00 g/ml, and the volume of the solvent (water) will be affected by the presence of NaCl, this conversion is not straightforward. However, assuming the volume change is negligible and the density is essentially that of water, we can proceed with an approximation.
Molality is defined as the number of moles of solute per kilogram of solvent. To find molality, we first need to determine the number of moles of NaCl in 1 liter of the solution, using the molarity (2.4 mol/L). Since the density is assumed to be 1.00 g/ml, 1 liter of solution would weigh 1000 g and thus contain 2.4 moles of NaCl.
Since the solution is mostly water, we will assume that the mass of the solvent is roughly 1000 g (1 kg), calculating on this basis would give us a molality of roughly 2.4 m (moles of NaCl per kg of water), since the additional mass of the NaCl is not significant enough to drastically change the mass of the water.
In reality, this is an approximation because the presence of NaCl does affect the mass and volume of the solution.