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SOLUBILITY
Definition: The Solubility of a solute (solid) in a solvent (liquid) is the concentration of the saturated solution. Solubility can be defined thus:
When some common salts is added to a beaker of water and the mixture stirred, the salt gradually disappear, and the clear colourless mixture is obtained. The salt is said to have dissolved in the water. Thus, the salt that dissolved in the water is called solute and the water that does the dissolving is known as the solvent and the product obtained by dissolving the salt in the water is called a solution.
Therefore, a solution is a homogenous mixture which is formed when a solute is completely dissolved in a solvent. A solution can be saturated or supersaturated.
Determination of solubility of KNO3 at 30oC
Stage 1: Preparing of Saturated Solution
Calculation of the solubility at 30oC
The following hypothetical values will be used to illustrate how to calculate the solubility of a salt.
Data required:
Solubility in mol.dm-3
Mass of anhydrous salt = (w – x) = (20.70 – 15.20) g = 5.50 g
i.e. 25 cm3 of saturated solution contain 5.50 g of salt,
Hence, 100 cm3 of saturated solution contain 5.50 x 1000/25 = 220 g of salt.
To convert 220 g to moles
Amount (mol) = (Mass of salt)/ (Molar mass).
Molar mass of KNO3 = 39 + 14 + 48 = 101 g.mol-1
i.e. Amount, in moles = 220/ 101 = 2.18 moles
Therefore, Solubility of KNO3 at 30oC = 2.18 mol.dm-3
Solubility in grams per 100 g of water
Mass of water used = (y – w) = (40.70 – 20.70 ) g = 20.0 g
Mass of salt used = (w – x) = (20.70 – 15.20) g = 5.50 g
i.e. 20.0 g of water at 30oC saturated 5.50 g of salt
Therefore, 100 g of water will saturate 5.50 x 100/20 g of salt
i.e. 5.50 x 5 g = 27.5 g of salt.
Hence, the solubility of the salt at 30oC is 27.5 g per 100 g of water.
WORKED EXAMPLES
Solution:
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