Solubility of a solute in a solvent at a particular temperature is the optimum amount of the solute in moles or grams that will saturate 1 dm3 or 1000cm3 of the solvent at a given temperature. Solubility is measured in mol/dm3.
Solubility is measured in mol/dm3 = mole of solute/mass of water (g) x 1000
Since, density of water = 1 g/cm3, 1g = 1cm3 for water.
Calculate the solubility of 4g of NaOH in 100g of water. (Na = 23, O = 16, H=1)
Mole = reacting mass in grams / Molar mass
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mole = 4/40 = 0.1 mole
Solubility in mol/dm3 = (0.1/100) x 100 = 1 mol/dm3
The solubility of sodium trioxocarbonate (iv) is exactly 2000 per 1000g of water at 90°c, and 500 per 1000g of water at 50°c. What will be the mass of sodium trioxocarbonate (iv) that will crystallize out of solution if 1000g of the saturated solution is at 90°c is cooled to 50°c.
Solute + solvent = solution
At 90°c, 2000g + 1000g = 3000g
At 50°c, 500g + 1000g = 1500g
Mass deposited on cooling from 900c to 500c
= 2000 – 500 = 1500g
3000g of saturated solution produces 1500g of solute.
1000g of the saturated solution would produce
= (1000 x 1500)/3000 = 500g of Na2CO3