Question

What is the molality of a solution prepared by dissolving 13.0 g of sucrose (C12H22O11) in 125 mL of a solvent with a density of 1.16 g/mL?

A. 0.30 m
B. 0.33 m
C. 0.038 m
D. 0.15 m
E. 0.26 m

Answer 1

To find the molality, calculate the mass of solvent in kg, find the moles of sucrose by dividing its mass by the molar mass, and then divide the moles by the mass of solvent in kg. The molality of the solution is 0.26 m.

Step-by-step explanation:

To calculate the molality of a solution prepared by dissolving 13.0 g of sucrose (C12H22O11) in 125 mL of a solvent with a density of 1.16 g/mL, follow these steps:

• First, calculate the mass of the solvent in kilograms. Since density is mass per volume, mass = (density) x (volume). So, the mass of 125 mL of a solvent with a density of 1.16 g/mL is 145 g, which is 0.145 kg.
• Next, calculate the molar mass of sucrose, which is approximately 342.3 g/mol.
• Then, determine the number of moles of sucrose by dividing the mass of sucrose by its molar mass. Thus, the moles of sucrose = (13.0 g) / (342.3 g/mol) = 0.038 mol.
• Now, you can find the molality by dividing the moles of sucrose by the kilograms of solvent. Molality = (0.038 mol) / (0.145 kg) = 0.26 m.