Molality $(m)$ of $3 \, \text{M}$ aqueous solution of $\text{NaCl}$ is: (Given: Density of solution $= 1.25 \, \text{g mL}^{-1}$, Molar mass in $\text{g mol}^{-1}$: $\text{Na} = 23, \, \text{Cl} = 35.5$)
To find the molality \(m\) of a 3 M aqueous solution of NaCl, we'll use the relationship between molality and molarity. Here is the step-by-step process:
Identify known values:
Molarity (M) = 3 M
Density of solution = 1.25 g/mL
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Calculate the mass of the solution: Since molarity (M) is given as 3 mol/L, it implies 3 moles of NaCl per 1 liter of solution.
Mass of 3 moles of NaCl = \(3 \times 58.5 = 175.5\) g
Total volume of the solution = 1000 mL (since we're considering 1 L for simplicity)
Mass of solution = Volume × Density = \(1000 \, \text{mL} \times 1.25 \, \text{g/mL} = 1250 \, \text{g}\)
Calculate the mass of water in the solution:
Mass of water = Mass of solution - Mass of NaCl
Mass of water = \(1250 - 175.5 = 1074.5 \, \text{g}\)
Convert mass of water to kg for molality: \(1074.5 \, \text{g} = 1.0745 \, \text{kg}\)
Calculate the molality: Molality \(m\) is given by the formula:
\(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\)
