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Difficulty: MediumSaturation States: Unsaturated, Saturated, and Supersaturated Solutions

A 200 cm3200\text{ cm}^3 sample of a saturated potassium trioxonitrate(V) solution, KNO3\text{KNO}_3, at 60∘C60^\circ\text{C} has a concentration of 1.5 mol/dm31.5\text{ mol/dm}^3. The solution is cooled to 25∘C25^\circ\text{C} to form a supersaturated solution. When a seed crystal is added, excess solute crystallizes until a new saturated concentration of 0.5 mol/dm30.5\text{ mol/dm}^3 is reached at 25∘C25^\circ\text{C}. What mass of KNO3\text{KNO}_3 crystallizes out of the solution? [Molar mass of KNO3=101 g/mol\text{KNO}_3 = 101\text{ g/mol}]

  1. A
    0.20 g0.20\text{ g}
  2. 20.2 g20.2\text{ g}Answer
  3. C
    30.3 g30.3\text{ g}
  4. D
    10.1 g10.1\text{ g}

Answer

20.2 g20.2\text{ g} of KNO3\text{KNO}_3 crystallizes out of the solution.
Cooling a saturated solution creates a unstable supersaturated state. Adding a seed crystal induces rapid crystallization of the excess solute until saturation equilibrium is re-established at the lower temperature. The difference in concentration is 1.0 mol/dm31.0\text{ mol/dm}^3. For 0.200 dm30.200\text{ dm}^3, this equals 0.20 mol0.20\text{ mol} of KNO3\text{KNO}_3, which has a mass of 0.20 mol×101 g/mol=20.2 g0.20\text{ mol} \times 101\text{ g/mol} = 20.2\text{ g}.

Step-by-Step Solution

1
Convert the volume of the solution from cm3\text{cm}^3 to dm3\text{dm}^3.
Volume=200 cm31000=0.200 dm3\text{Volume} = \frac{200\text{ cm}^3}{1000} = 0.200\text{ dm}^3.
Concentrations are given in mol/dm3\text{mol/dm}^3, so volume must be in dm3\text{dm}^3.
2
Calculate the moles of KNO3\text{KNO}_3 dissolved initially at 60∘C60^\circ\text{C} and remaining at 25∘C25^\circ\text{C}.
Initial moles=1.5 mol/dm3×0.200 dm3=0.30 mol\text{Initial moles} = 1.5\text{ mol/dm}^3 \times 0.200\text{ dm}^3 = 0.30\text{ mol}. Final moles=0.5 mol/dm3×0.200 dm3=0.10 mol\text{Final moles} = 0.5\text{ mol/dm}^3 \times 0.200\text{ dm}^3 = 0.10\text{ mol}.
Determining the mole difference shows how much solute leaves the supersaturated state upon seeding.
3
Determine the amount of moles precipitated and convert to mass.
Moles precipitated=0.30 mol−0.10 mol=0.20 mol\text{Moles precipitated} = 0.30\text{ mol} - 0.10\text{ mol} = 0.20\text{ mol}. Mass precipitated=0.20 mol×101 g/mol=20.2 g\text{Mass precipitated} = 0.20\text{ mol} \times 101\text{ g/mol} = 20.2\text{ g}.
Multiplying the precipitated moles by the molar mass gives the required mass in grams.

Key Concept

Mass of solute crystallized from a supersaturated solution upon reaching saturation equilibrium
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Topics

Saturation States: Unsaturated, Saturated, and Supersaturated Solutions