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

The solubility of potassium trioxonitrate(V), KNO3\text{KNO}_3, in water at 40∘C40^\circ\text{C} is 6.0 mol dm−36.0\text{ mol dm}^{-3}. What mass of KNO3\text{KNO}_3, in grams, is required to prepare a saturated solution in 250 cm3250\text{ cm}^3 of water at this temperature? (Molar mass of KNO3=101 g mol−1\text{KNO}_3 = 101\text{ g mol}^{-1})

Answer: 151.5 g

Answer

The mass of KNO3\text{KNO}_3 required to prepare a saturated solution in 250 cm3250\text{ cm}^3 of water at 40∘C40^\circ\text{C} is 151.5 g151.5\text{ g}.
To find the mass of solute required for saturation, convert the given volume of water to cubic decimeters (250 cm3=0.25 dm3250\text{ cm}^3 = 0.25\text{ dm}^3). Multiply the volume by the molar solubility (6.0 mol dm−3×0.25 dm3=1.5 mol6.0\text{ mol dm}^{-3} \times 0.25\text{ dm}^3 = 1.5\text{ mol}) to obtain the number of moles, then multiply by the molar mass (1.5 mol×101 g mol−1=151.5 g1.5\text{ mol} \times 101\text{ g mol}^{-1} = 151.5\text{ g}).

Step-by-Step Solution

1
Convert volume from cm3\text{cm}^3 to dm3\text{dm}^3
0.25 dm30.25\text{ dm}^3
Molar solubility is expressed per dm3\text{dm}^3, so the volume of solvent must be in dm3\text{dm}^3.
2
Determine moles of KNO3\text{KNO}_3 needed for saturation
1.5 mol1.5\text{ mol}
Multiply molar solubility by the volume in dm3\text{dm}^3.
3
Convert moles to mass in grams
151.5 g151.5\text{ g}
Multiply moles by the molar mass of KNO3\text{KNO}_3 (101 g mol−1101\text{ g mol}^{-1}).

Key Concept

Calculating solute mass for saturation using molar solubility and volume
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Topics

Solubility, Solubility Curves, and Solubility Product (Ksp)

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