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What is the molality of a solution prepared by dissolving 2.00 ml of bromine (d = 3.105 g/ml) in 125 ml of acetic acid, hc2h3o2 (d = 1.05 g/ml) (ans. 0.30 m)?

Respuesta :

The molality of a solution is 0.304g

Calculation,

Given data,

  • Volume of bromine = 2 ml
  • density of bromine = 3.105 g/ml
  • Volume of acetic acid = 125 ml
  • density of of acetic acid = 1.05 g/ml

Density of bromine = 3.105 g/ml  = mass / volume = mass / 2 ml

Mass of bromine = 3.105 g/ml ×  2 ml = 6.206 g

Density of acetic acid = 1.05 g/ml = mass / volume = mass / 125 ml

Mass of bromine = 1.05 g/ml ×  125 ml = 127.5 g

Molality = mass of solute in gram/molar mass of solute × 1000/mass of solvent in gram

Molality = 6.206 g/159.88 g/mol × 1000/127.5 g = 0.304 g

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