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Determine the molality and molarity of a sodium chloride solution prepared by adding 52.80 g of solid NaCl to 150.0 g of water. The density of the saline solution is 1.1972 g mL . Molality

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Answer:

  • 6.02 m
  • 5.334 M

Explanation:

First we convert 52.80 g of NaCl to moles, using its molar mass:

  • 52.80 g ÷ 58.44 g/mol = 0.9035 mol

To determine the molality, we convert 150.0 g of water to kg:

  • 150.0 g / 1000 = 0.150 kg
  • molality = 0.9035 mol / 0.150 kg =  6.02 m

As for the molarity, we determine the total mass of solution:

  • 52.80 g + 150.0 g = 202.8 g

And calculate the volume, using the density:

  • 202.8 g ÷ 1.1972 g/mL = 169.39 mL
  • 169.39 mL / 1000 = 0.16939 L

Finally we calculate the molarity:

  • 0.9035 mol / 0.16939 L = 5.334 M

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