A student dissolves 12. g of sucrose C12H22O11 in 300. mL of a solvent with a density of 1.01/gmL. The student notices that the volume of the solvent does not change when the sucrose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits. molarity = molality =

Molarity → 0.17 M

Molality → 0.11 m

Explanation:

The student notices that the volume of the solvent does not change when the sucrose dissolves in it; therefore we assume the volume of solvent as solution.

Molarity = Mol of solute/L

Let's calculate the mol of solute (mass / molar mass)

12 g / 342 g/mol = 0.0351 moles

Let's conver the volume (mL) to L

300 mL / 1000 = 0.3 L

Molarity (mol/L) = 0.0351 mol / 0.3L → 0.17 M

Molality = mol of solute / 1kg of solvent.

Let's find out the mass of solvent with the density

Solvent density = Solvent mass / Solvent volume

1.01 g/mL = Solvent mass / 300 mL

1.01 g/mL. 300 mL = Solvent mass →303 g

Let's convert the mass to kg

303 g / 1000 = 0.303 kg

Molality (mol/kg) → 0.0351 mol / 0.303kg = 0.11 m