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Difficulty: HardSolubility, Solubility Curves, and Solubility Product (Ksp)

At 60∘C60^\circ\text{C}, the solubility of a trioxonitrate(V) salt YY (molar mass =160 g mol−1= 160\text{ g mol}^{-1}) is 1.25 mol dm−31.25\text{ mol dm}^{-3}. When the solution is cooled to 20∘C20^\circ\text{C}, its solubility decreases to 0.25 mol dm−30.25\text{ mol dm}^{-3}. What mass of salt YY will crystallize out when 300 g300\text{ g} of a saturated solution of YY at 60∘C60^\circ\text{C} is cooled to 20∘C20^\circ\text{C}? (Assume the density of water is 1.0 g cm−31.0\text{ g cm}^{-3}).

  1. 40.0 g40.0\text{ g}Answer
  2. B
    48.0 g48.0\text{ g}
  3. C
    50.0 g50.0\text{ g}
  4. D
    160.0 g160.0\text{ g}

Answer

The mass of salt YY that crystallizes out upon cooling is 40.0 g40.0\text{ g}.
Converting concentrations to mass per 1000 g1000\text{ g} of water gives 200 g200\text{ g} at 60∘C60^\circ\text{C} and 40 g40\text{ g} at 20∘C20^\circ\text{C}. A 300 g300\text{ g} sample of saturated solution at 60∘C60^\circ\text{C} contains 50 g50\text{ g} of solute dissolved in 250 g250\text{ g} of solvent (water). Upon cooling to 20∘C20^\circ\text{C}, 250 g250\text{ g} of water can only hold 10 g10\text{ g} of solute. Thus, 50 g−10 g=40.0 g50\text{ g} - 10\text{ g} = 40.0\text{ g} crystallizes out.

Step-by-Step Solution

1
Convert solubility at 60∘C60^\circ\text{C} from mol dm−3\text{mol dm}^{-3} to g dm−3\text{g dm}^{-3}
Solubility at 60∘C=1.25 mol dm−3×160 g mol−1=200 g dm−3\text{Solubility at } 60^\circ\text{C} = 1.25\text{ mol dm}^{-3} \times 160\text{ g mol}^{-1} = 200\text{ g dm}^{-3}
Solubility calculations involving mass of solution require mass concentration (g dm−3\text{g dm}^{-3}).
2
Determine mass of water in 300 g300\text{ g} of saturated solution at 60∘C60^\circ\text{C}
Mass of solution per dm3=1000 g (water)+200 g (salt)=1200 g\text{Mass of solution per dm}^3 = 1000\text{ g (water)} + 200\text{ g (salt)} = 1200\text{ g}. Therefore, mass of water in 300 g solution=1000 g1200 g×300 g=250 g\text{mass of water in } 300\text{ g solution} = \frac{1000\text{ g}}{1200\text{ g}} \times 300\text{ g} = 250\text{ g}.
Solubility expresses mass of solute per fixed mass of solvent (1000 g1000\text{ g} water).
3
Convert solubility at 20∘C20^\circ\text{C} to g dm−3\text{g dm}^{-3} and find mass of salt remaining dissolved in 250 g250\text{ g} water
Solubility at 20∘C=0.25 mol dm−3×160 g mol−1=40 g dm−3\text{Solubility at } 20^\circ\text{C} = 0.25\text{ mol dm}^{-3} \times 160\text{ g mol}^{-1} = 40\text{ g dm}^{-3}. In 250 g250\text{ g} of water, dissolved salt =40 g×250 g1000 g=10 g= 40\text{ g} \times \frac{250\text{ g}}{1000\text{ g}} = 10\text{ g}.
Determining how much solute stays dissolved at the lower temperature.
4
Calculate the mass of salt precipitated
Original salt in solution=200 g×250 g1000 g=50 g\text{Original salt in solution} = 200\text{ g} \times \frac{250\text{ g}}{1000\text{ g}} = 50\text{ g}. Mass precipitated=50 g−10 g=40.0 g\text{Mass precipitated} = 50\text{ g} - 10\text{ g} = 40.0\text{ g}.
Mass precipitated is the difference between initial dissolved mass and remaining dissolved mass.

Key Concept

Mass of salt precipitated upon cooling saturated solutions
Estimated Time:2m 30s
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Topics

Solubility, Solubility Curves, and Solubility Product (Ksp)

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