Concentration of Aqueous Phosphoric Acid Solution

The mole fraction of phosphoric acid in the tank at 25 minutes can be calculated based on the given data:

Calculation of Mole Fraction of Phosphoric Acid

Given data:
Initial moles of phosphoric acid in the tank = 150 kmol
Initial mole fraction of H3PO4 = 5.00% = 0.05
Rate of addition of pure phosphoric acid = 30.0 L/min
Molecular weight of phosphoric acid = 98
Density of phosphoric acid = 1.834 kg/L

Calculation:
- Moles of phosphoric acid added in 25 minutes = 30 L/min * 25 min = 750 L
- Mass of phosphoric acid added = Density * Volume added = 1.834 kg/L * 750 L = 1375.5 kg = 1375.5/98 = 14.036 kmol
- Moles of phosphoric acid in the tank after addition = Initial moles + Moles added = 150 kmol * 0.05 + 14.036 kmol = 21.536 kmol
- Moles of water in the tank after addition = Initial moles * (1 - Initial mole fraction) = 150 kmol * 0.95 = 142.5 kmol

Now, calculate the mole fraction of phosphoric acid in the tank:
Mole fraction = Moles of phosphoric acid / (Moles of phosphoric acid + Moles of water)
Mole fraction = 21.536 kmol / (21.536 kmol + 142.5 kmol) = 0.1313

Therefore, the mole fraction of phosphoric acid in the tank at 25 minutes is 0.1313.