Example 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,)
Solution
30cm 3 of saturated solution has 2.5g
1,000cm 3. in the mixture will be able to contain
1000cm330cm3x2.5g1=83.3g/dm3
Solubility in mols/dm3 = Solubilitying/dm3MolarMass
Molar mass Nax = 19 + 23 = 42
= 83.342 = 1.98mol/dm3
Or
Solubility = MassofsoluteVolumeofsolventx1000MolarMass
= 2.530×100042
= 1.98mol/dm3
Example 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?
- Moles per dm3 [K=39, Cl=35.5]
Solution
(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
Example 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.
Solution
At 32degc
| Mass of the dish that evaporates + CuSO 4 solid | 33.98g | |
| Mass of dish that evaporates | – | 31.18g |
| Volume of CuSO 4(s) solid | = | 2.80g |
| Mass of the dish used to evaporate + CuSO 4(aq) solution | 48.39g | |
| Mass of dish for evaporation + CuSO 4(s) solid | – | 33.98g |
| The amount of solvent | = | 14.41g |
14.41g of the solvent were saturated with 2.80g of CuSO 4(s) solid
1g solvent is saturated with 2.8014.41 of CuSO 4(s)solid
The solvent contained in 100g was saturated with
2.8014.41x1001ofCuSO4(s)
= 19.43g/100g of solvent
Example 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 =
Massofsolutevol/massofsolventx10001
= 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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