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The solubility of a compound with a molecular weight of 233.2 is determined to be 27.9 g in 125.0 g of water at 398 K. Express this concentration in terms of molality. 0.957

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Answer:

0.960 m

Explanation:

Given data

  • Mass of the solute: 27.9 g
  • Molar mass of the solute: 233.2 g/mol
  • Mass of the solvent: 125.0 g = 0.1250 kg

First, we will calculate the moles of solute.

27.9 g × (1 mol/233.2 g) = 0.120 mol

The molality of the compound is:

m = moles of solute / kilograms of solvent

m = 0.120 mol / 0.1250 kg

m = 0.960 m

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