Questions: A student dissolves 6.7 g of sucrose (C12H22O11) in 200 mL of a solvent with a density of 0.91 g/mL. 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.

A student dissolves 6.7 g of sucrose (C12H22O11) in 200 mL of a solvent with a density of 0.91 g/mL. 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.
Transcript text: A student dissolves 6.7 g of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ in $200 . \mathrm{mL}$ of a solvent with a density of $0.91 \mathrm{~g} / \mathrm{mL}$. 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.
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Solution

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Solution Steps

Step 1: Calculate the moles of sucrose

The molar mass of sucrose (C12H22O11) is 342.3 g/mol. We can calculate the number of moles by dividing the mass of sucrose by its molar mass:

Moles of sucrose = (6.7 g) / (342.3 g/mol) = 0.0196 mol

Step 2: Calculate the molarity

Molarity is defined as moles of solute per liter of solution. The volume of the solvent is 200 mL, which is equal to 0.200 L. Assuming the volume doesn't change upon dissolving the sucrose, the molarity is:

Molarity = (0.0196 mol) / (0.200 L) = 0.098 M

Step 3: Calculate the mass of the solvent

The mass of the solvent can be calculated using its volume and density:

Mass of solvent = (200. mL) * (0.91 g/mL) = 182 g

Convert this to kilograms: 182g = 0.182 kg

Step 4: Calculate the molality

Molality is defined as moles of solute per kilogram of solvent. We use the calculated moles from Step 1 and the mass of the solvent in kilograms:

Molality = (0.0196 mol) / (0.182 kg) = 0.11 mol/kg

Final Answer:

The molarity of the solution is 0.098 M, and the molality of the solution is 0.11 mol/kg.

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