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What is the molality of a solution prepared by dissolving 0.500 moles of CaF₂ in 11.5 moles of H₂O?

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Final answer:

The molality of the solution is approximately 2.41 m, calculated by dividing 0.500 moles of CaF₂ by the mass of the water in kilograms.

Step-by-step explanation:

The molality of a solution is defined as the number of moles of solute per kilogram of solvent. To calculate the molality of the solution prepared by dissolving 0.500 moles of CaF₂ in 11.5 moles of H₂O, we first need to convert the amount of water to kilograms. Since the molar mass of water (H₂O) is approximately 18.015 g/mol, we can calculate the mass of the solvent.

First, convert the moles of H₂O to grams: (11.5 moles H₂O) × (18.015 g/mol H₂O) = 207.1725 g H₂O

Convert grams to kilograms: 207.1725 g H₂O × (1 kg / 1000 g) = 0.2071725 kg H₂O

Finally, calculate the molality (m):
m = moles of solute / kg of solvent = 0.500 moles CaF₂ / 0.2071725 kg H₂O ≈ 2.41 m (molality)

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