Answer:
0.1025 g/L
Explanation:
From the question above,
The solubility of oxygen gas in water at 25°c and 1.0 atm pressure is 0.041g/L
Henry's law states at a constant temperature the solubility of a gas is proportional to the pressure
Since the temperatures are both similar (25°c) then, the solubility of oxygen in water at 2.5 atm can be calculated as follows
= 0.041/1.0 × 2.5
= 0.041×2.5
= 0.1025 g/L
Hence the solubility of oxygen in water at 2.5 atm and 25°c is 0.1025 g/L
