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using 680g of sodium sulphate anhydrous what is the increase of sodium in 600 litres water
To calculate the increase in sodium concentration, we first need to determine the amount of sodium in the added sodium sulfate.
1. Calculate the moles of sodium sulfate:
Molar mass of Na2SO4 = (2 * 22.99 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol) = 142.04 g/mol
Moles of Na2SO4 = Mass / Molar mass = 680 g / 142.04 g/mol ≈ 4.79 mol
2. Calculate the moles of sodium ions:
Each mole of Na2SO4 produces 2 moles of Na+ ions.
Moles of Na+ = 2 * Moles of Na2SO4 = 2 * 4.79 mol ≈ 9.58 mol
3. Calculate the mass of sodium added:
Mass of Na+ = Moles of Na+ * Molar mass of Na = 9.58 mol * 22.99 g/mol ≈ 220.4 g
4. Calculate the concentration of sodium in the solution:
Volume of solution = 600 L = 600,000 mL
Concentration of Na+ = Mass of Na+ / Volume of solution = 220.4 g / 600,000 mL ≈ 0.000367 g/mL
5. Convert the concentration to a more common unit (mg/L):
Concentration of Na+ = 0.000367 g/mL * 1000 mg/g * 1000 mL/L ≈ 367 mg/L
Therefore, adding 680 g of sodium sulfate anhydrous to 600 liters of water will increase the sodium concentration by approximately 367 mg/L.