Final answer:
The molality of a 3.1416 M sucrose solution with a density of 1.5986 g/mL is calculated by finding the mass of sucrose and water in one liter, then dividing the moles of sucrose by the kilograms of water.
Step-by-step explanation:
To calculate the molality of a given 3.1416 M aqueous solution of sucrose (C12H22O11) with a density of 1.5986 g/mL, first, we need to find the mass of sucrose and the mass of water in one liter of the solution. Using the molar mass of sucrose, which is 342.297 g/mol, we find that one liter of solution contains 3.1416 moles of sucrose, or 1075.4 grams (3.1416 mol × 342.297 g/mol).
We then convert the density to grams per liter by multiplying by 1000 mL, giving us 1598.6 grams of the solution in one liter (1.5986 g/mL × 1000 mL/L). Subtracting the mass of sucrose from this amount, we have 523.2 grams of water left (1598.6 g - 1075.4 g).
Converting the mass of water to kilograms results in 0.5232 kg. Lastly, we divide the moles of sucrose by the kilograms of water to find the molality: 3.1416 mol sucrose / 0.5232 kg water, which gives us the final molality of the solution.