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Zorluk: ZorSolubility Curves and Temperature Effects

The solubility of a salt ZZ (molar mass = 101.0 g mol−1101.0\text{ g mol}^{-1}) in water is 5.0 mol dm−35.0\text{ mol dm}^{-3} at 80∘C80^\circ\text{C} and 202.0 g dm−3202.0\text{ g dm}^{-3} at 25∘C25^\circ\text{C}. If 250 cm3250\text{ cm}^3 of a saturated solution of salt ZZ is cooled from 80∘C80^\circ\text{C} to 25∘C25^\circ\text{C}, what mass of salt ZZ will crystallize out of the solution?

  1. 75.75 g75.75\text{ g}Cevap
  2. B
    303.00 g303.00\text{ g}
  3. C
    50.50 g50.50\text{ g}
  4. D
    0.75 g0.75\text{ g}

Cevap

The mass of salt Z that will crystallize out of solution is 75.75 g75.75\text{ g}.
Converting the solubility at 80∘C80^\circ\text{C} (5.0 mol dm−35.0\text{ mol dm}^{-3}) into mass concentration yields 505.0 g dm−3505.0\text{ g dm}^{-3}. Subtracting the solubility at 25∘C25^\circ\text{C} (202.0 g dm−3202.0\text{ g dm}^{-3}) gives 303.0 g dm−3303.0\text{ g dm}^{-3} precipitated. Multiplying by the volume ratio (250 cm3/1000 cm3=0.25250\text{ cm}^3 / 1000\text{ cm}^3 = 0.25) yields 75.75 g75.75\text{ g}.

Adım Adım Çözüm

1
Convert the solubility at 80∘C80^\circ\text{C} from mol dm−3\text{mol dm}^{-3} to g dm−3\text{g dm}^{-3}.
Solubility at 80∘C=5.0 mol dm−3×101.0 g mol−1=505.0 g dm−3\text{Solubility at } 80^\circ\text{C} = 5.0\text{ mol dm}^{-3} \times 101.0\text{ g mol}^{-1} = 505.0\text{ g dm}^{-3}.
Solubility values given in molarity must be multiplied by molar mass to obtain concentration in mass per unit volume.
2
Calculate the mass of solute precipitated per dm3\text{dm}^3 upon cooling to 25∘C25^\circ\text{C}.
ΔSolubility=505.0 g dm−3−202.0 g dm−3=303.0 g dm−3\Delta\text{Solubility} = 505.0\text{ g dm}^{-3} - 202.0\text{ g dm}^{-3} = 303.0\text{ g dm}^{-3}.
The difference between solubilities at the higher and lower temperatures gives the mass of solute that cannot remain dissolved in 1 dm31\text{ dm}^3 of water.
3
Scale the precipitated mass to the given volume of 250 cm3250\text{ cm}^3.
Mass crystallized=303.0 g dm−3×250 cm31000 cm3 dm−3=303.0 g×0.25=75.75 g\text{Mass crystallized} = 303.0\text{ g dm}^{-3} \times \frac{250\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 303.0\text{ g} \times 0.25 = 75.75\text{ g}.
The solution volume is 250 cm3250\text{ cm}^3 (0.25 dm30.25\text{ dm}^3), so the precipitated mass is proportional to this fractional volume.

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