The concentration of oxygen in water at the bottom of a lake is 0.48 g/L and the pressure is 2.5 atm. If water from the bottom is moved by a current upwards to a depth where the pressure is 1.3 atm, what is the concentration of the oxygen in the water at this depth?

[tex]P = \frac{nRT}{m/ \rho} \\\\P = \frac{nRT \rho}{m} \\\\P = k\rho\\\\k = \frac{P_1}{\rho _1} = \frac{P_2}{\rho _2} \\\\\rho_2 = \frac{P_2\rho_1}{P_1} \\\\\rho_2 = \frac{1.3 \times 0.48}{2.5} \\\\\rho_2 = 0.25 \ g/L[/tex]

Thus, the concentration of the oxygen in the water at this depth is 0.25 g/L.

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shristiparmar1221 shristiparmar1221 · Answer 2 · 2022-06-21T20:34:51+02:00

The concentration of the oxygen in the water at this depth is 0.25 g/L If water from the bottom is moved by a current upwards to a depth where the pressure is 1.3 atm.

What is Concentration ?

The concentration of a chemical substance expresses the amount of a substance present in a mixture.

There are many different ways to express concentration.

Let's Apply Ideal gas law to obtain concentration of the oxygen ;

PV = nRT

P = nRT/V

P = Кр

K = P₁ / P¹ = P₂ / P²

P₂ =P₁ P² / P¹

P₂ = 1.3 x 0.48 / 2.5

P₂ = 0.25 g/l

Thus, the concentration of the oxygen in the water at this depth is 0.25 g/L.

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