Wéi Berechent Solubilitéit

Wéi Berechent Solubilitéit

Beispill 1

30cm 3 of an insoluble solution of salt NaX was dissolved in water at 25°C and evaporated until dryness, and only 2.5g of remaining solid material was left. Determine the amount of salt solubilized in the water 25°C as moles per dm 3.(Na = 23, X = 19,)


30cm 3 of saturated solution has 2.5g

1,000cm 3. in the mixture will be zouverléisseg enthalen


Solubility in mols/dm3 = Solubilitying/dm3MolarMass

Molar mass Nax = 19 + 23 = 42

= 83.342 = 1.98mol/dm3


Solubility = MassofsoluteVolumeofsolventx1000MolarMass

= 2.530 × 100042

= 1.98mol/dm3

Beispill 2

At 25°C, 23g of KCl dissolves into 100 centimeters 3.0 of water, resulting in an aqueous solution. What is the salt’s solubility in 1. Grammes for every dm 3. of the solvent at this temperature?

  1. Moles per dm3 [K=39, Cl=35.5]


(1) Solubility in g/dm3 = MassofsoluteVolumeofsolventx10001

= 23g100cm3x1000cm31

= 230g/dm3

(2) RMM of KCl = 39+35.5 = 74.5

Solubility in mol/dm3 = Solubilityingdm-3Molarmass

= 23074.5

= 3.09mol/dm3

Beispill 3

An evaporating dish weighed 31.18g, when filled with saturated solution of copper teteraoxosulphate(vi) solution at 32degC it weighed 48.39g, after evaporation to dryness the evaporating dish and residue weighed 33.98g.

Determine the concentration of solubility for copper (II) Tetraoxosulphate (IV) crystals at temperatures of 32 degrees Celsius in grams in 100g of the solvent.


At 32degc

Mass of the dish that evaporates + CuSO 4 staark33.98g
Mass of dish that evaporates-31.18g
Volume of CuSO 4 (s) staark=2.80g
Mass of the dish used to evaporate + CuSO 4 (aq) Léisung48.39g
Mass of dish for evaporation + CuSO 4 (s) staark-33.98g
The amount of solvent=14.41g
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14.41g of the solvent were saturated with 2.80g of CuSO 4 (s) staark

1g solvent is saturated with 2.8014.41 of CuSO 4 (s)staark

The solvent contained in 100g was saturated with


= 19.43g/100g of solvent

Beispill 4

250cm 3 of saturated solution of Sodium Chloride has 5.85g of salt that has been that is dissolved within it at the temperature of room. How much is the solubility salt? (i) grams per dm 3. (ii) mole/dm 3 [Na = 23 Cl =35.5]

Solubility in g/dm 3 =


= 5.85g250cm3x10001

= 23.4g/dm3

Solubility in mol/dm3 = SolubilityingMolarMass

Molar mass NaCl is 23 + 35.5 = 58.5

= 23.458.5

= 0.4mol/dm3


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