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Difficulty: HardSolubility Curves and Temperature Effects

The solubility of potassium trioxonitrate(V), KNO3\text{KNO}_3, in water is 3.5 mol dm−33.5\text{ mol dm}^{-3} at 70∘C70^\circ\text{C} and 1.2 mol dm−31.2\text{ mol dm}^{-3} at 25∘C25^\circ\text{C}. What mass of KNO3\text{KNO}_3 will crystallize out of solution when 200 cm3200\text{ cm}^3 of a saturated solution at 70∘C70^\circ\text{C} is cooled to 25∘C25^\circ\text{C}? [K=39,N=14,O=16][\text{K} = 39, \text{N} = 14, \text{O} = 16]

  1. 46.46 g46.46\text{ g}Answer
  2. B
    0.46 g0.46\text{ g}
  3. C
    70.70 g70.70\text{ g}
  4. D
    24.24 g24.24\text{ g}

Answer

The mass of potassium trioxonitrate(V) that crystallizes out is 46.46 g46.46\text{ g}.
The net solubility decrease when cooling from 70∘C70^\circ\text{C} to 25∘C25^\circ\text{C} is 2.3 mol dm−32.3\text{ mol dm}^{-3}. In 200 cm3200\text{ cm}^3 (0.2 dm30.2\text{ dm}^3) of saturated solution, 0.46 mol0.46\text{ mol} of KNO3\text{KNO}_3 precipitates out. Multiplying 0.46 mol0.46\text{ mol} by the molar mass of KNO3\text{KNO}_3 (101 g mol−1101\text{ g mol}^{-1}) yields 46.46 g46.46\text{ g}.

Step-by-Step Solution

1
Calculate the molar mass of potassium trioxonitrate(V), KNO3\text{KNO}_3.
Molar mass=39+14+(3×16)=101 g mol−1\text{Molar mass} = 39 + 14 + (3 \times 16) = 101\text{ g mol}^{-1}.
Molar mass is needed to convert molar concentration into mass.
2
Find the change in solubility per dm3\text{dm}^3 upon cooling from 70∘C70^\circ\text{C} to 25∘C25^\circ\text{C}.
ΔS=3.5 mol dm−3−1.2 mol dm−3=2.3 mol dm−3\Delta S = 3.5\text{ mol dm}^{-3} - 1.2\text{ mol dm}^{-3} = 2.3\text{ mol dm}^{-3}.
The crystallization amount depends on the difference between initial and final solubilities.
3
Determine the amount of solute precipitated in 200 cm3200\text{ cm}^3 (0.2 dm30.2\text{ dm}^3) of solution.
Moles=2.3 mol dm−3×0.2 dm3=0.46 mol\text{Moles} = 2.3\text{ mol dm}^{-3} \times 0.2\text{ dm}^3 = 0.46\text{ mol}.
Solubility values are given per dm3\text{dm}^3, so they must be scaled to the given volume of 200 cm3200\text{ cm}^3.
4
Convert the precipitated moles into mass in grams.
Mass=0.46 mol×101 g mol−1=46.46 g\text{Mass} = 0.46\text{ mol} \times 101\text{ g mol}^{-1} = 46.46\text{ g}.
Multiplying the number of moles by molar mass gives the required mass in grams.

Key Concept

Solubility Curves and Temperature Effects
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Topics

Solubility Curves and Temperature Effects

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