If \(100^3\) of a saturated solution of sodium tetraoxosulphate (VI) at 30° C contains 10.5 g of the salt, what would be its solubility at this temperature? [(\(Na_2SO_4\) = 142]
Molar mass of Na2SO4 = 2(Na) + S + 4(O) = 2(23.0 g/mol) + 32.1 g/mol + 4(16.0 g/mol) = 46.0 g/mol + 32.1 g/mol + 64.0 g/mol = 142.1 g/mol
Now, we can calculate the number of moles of Na2SO4 in the solution:
Number of moles = Mass of the salt / Molar mass
Number of moles = 10.5 g / 142.1 g/mol ≈ 0.0739 mol
Next, we need to find the volume of the saturated solution, given as 100 cm³, and convert it to liters:
Volume of solution = 100 cm³ = 100 cm³ / 1000 cm³/dm³ = 0.1 dm³
Finally, we can calculate the solubility in mol/dm³:
Solubility = Number of moles / Volume of solution
Solubility = 0.0739 mol / 0.1 dm³ ≈ 0.739 mol/dm³
Rounded to two decimal places, the solubility of sodium tetraoxosulphate (VI) at 30°C is approximately 0.74 mol/dm³.
Therefore, the correct option is:
0.74 mol dm-3.
Solubility = 10.5 g / 100 cm3 * 142 g/mol = 0.739 mol/dm3
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