Density of 3 M NaCl solution is $1.25 \mathrm{~g} / \mathrm{mL}$. The molality of the solution is :

<p>$\text{Molarity of NaCl} = 3 \text{ M} \quad (\text{3 moles in 1 L of solution})$</p> <p>Given the density of the solution is $1.25 \, \text{g/mL},$ the total mass of 1 liter (1000 mL) of the solution is:</p> <p>$1.25 \, \text{g/mL} \times 1000 \, \text{mL} = 1250 \, \text{g}$</p> <p>To find the molality, we need the mass of the solvent (water). First, calculate the mass of NaCl in 1 L of solution. The molar mass of NaCl is approximately:</p> <p>$$ 23 \, \text{g/mol (Na)} + 35.45 \, \text{g/mol (Cl)} \approx 58.45 \, \text{g/mol} $$</p> <p>Thus, the mass of NaCl is:</p> <p>$3 \, \text{mol} \times 58.45 \, \text{g/mol} \approx 175.35 \, \text{g}$</p> <p>Now, the mass of the solvent (water) is:</p> <p>$$ 1250 \, \text{g (solution)} - 175.35 \, \text{g (NaCl)} \approx 1074.65 \, \text{g} $$</p> <p>Convert the mass of the solvent to kilograms:</p> <p>$1074.65 \, \text{g} \div 1000 \approx 1.07465 \, \text{kg}$</p> <p>Molality ($m$) is defined as the number of moles of solute per kilogram of solvent:</p> <p>$m = \frac{3 \, \text{mol}}{1.07465 \, \text{kg}} \approx 2.79 \, \text{mol/kg}$</p> <p>Thus, the molality of the solution is approximately $2.79 \, \text{m}.$</p>