If \(100^3\) of a saturated solution of sodium tetraoxosulphate (VI) at 30° C contains 10.5
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]
- A. 2.15 mol dm-3
- B. 0.74 mol dm-3
- C. 0.60 mol dm-3
- D. 0.57 mol dm-3
Correct Answer: B. 0.74 mol dm-3
Explanation
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