Calculate the molarity of 139 grams of sucrose, [tex]C_{12}H_{22}O_{11}[/tex], in 2.60 liters of solution.

The molality of a 3.1416 M aqueous solution of sucrose with a density of 1.5986 is calculated to be approximately 6.005 mol/kg, using steps that include finding the total mass of the solution, the mass of sucrose, and then the mass of water to determine the molality.

Explanation:

To calculate the molality of a 3.1416 M aqueous solution of sucrose with a density of 1.5986, we first convert the density to grams per liter and then find the mass of the solute and the solvent. The molar mass of sucrose, C₁₂H₂₂O₁₁, is 342.297 g/mol. To begin, we would calculate the total mass of the solution per liter:

1.5986 g/mL × 1000 mL/L = 1598.6 g/L solution

Then, knowing the molarity, we can find the mass of sucrose per liter of solution:

3.1416 mol/L × 342.297 g/mol = 1075.4 g sucrose/L solution

We subtract the mass of sucrose from the total mass of the solution to find the mass of water:

1598.6 g solution - 1075.4 g sucrose = 523.2 g water

Convert this mass to kilograms (since molality requires kilograms of solvent):

523.2 g water × 1 kg/1000 g = 0.5232 kg water

Finally, we calculate the molality (moles of solute per kilogram of solvent):

3.1416 mol sucrose / 0.5232 kg water = 6.005 mol/kg