We are given: - Molarity (M) = 3 M (mol/L) - Density of solution = 1.25 g/mL = 1250 g/L
The molality (m) is given by the formula: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
From molarity, the number of moles of NaCl in 1 liter of solution is: \[ \text{moles of NaCl} = 3 \, \text{mol} \]
To find the mass of NaCl, use the molar mass of NaCl (58.44 g/mol): \[ \text{mass of NaCl} = 3 \, \text{mol} \times 58.44 \, \text{g/mol} = 175.32 \, \text{g} \]
Now, the mass of the solvent (water) is: \[ \text{mass of solvent} = 1250 \, \text{g} - 175.32 \, \text{g} = 1074.68 \, \text{g} = 1.07468 \, \text{kg} \]
Thus, the molality is: \[ m = \frac{3 \, \text{mol}}{1.07468 \, \text{kg}} = 2.79 \, \text{m} \]
Therefore, the correct answer is 2.79 m.