calculates the density (g / ml) and the molality of an aqueous solution of HF at...

Question

Question

calculates the density (g / ml) and the molality of an aqueous solution of HF at 48.0% by mass whose concentration is 27.60M

Answer 1

Answer #1

HF at 48.0% by mass means 48.0 g HF remain in 100 g solution.

mass of solvent = (100 - 48) = 52.0 g = 0.052 kg.

48.0 g HF = mass / molar mass = 48.0 g / 20.01 g/mole = 2.399 mole.

thus

molality = 2.399 mole / 0.052 kg = 46.13 m

concentration is 27.60 M means 27.60 mole HF remain in 1000 ml solution.

27.60 mole HF = mole * molar mass = 27.60 mole * 20.01 g / mole = 552.276 g

thus

density = mass / volume = 552.276 g / 1000 ml = 0.552 g / ml

Answer 2

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