Question

The amount of oxygen dissolved in 1dm3 of water at 293 k will be? Henry constant is 4.58

Answer 1

Answer:

Explanation:

Given The amount of oxygen dissolved in 1 dm^3 of water at 293 k will be? Henry constant is 4.58

We know that

 X2 = p/Kh

      = 0.2 / 4.58 x 10^4

      = 4.35 x 10^-4

Now X2 = n2 / n1 + n2

             = n2 / 18 x 10^-3 + n2

      4.35 x 10^-6 = n2/18 x 10^-3 + n2

n 2 = 2.41 x 10^-4

now the amount of oxygen dissolved will be 2.41 x 10^-4 x 32 x 10^-3

                                                                    = 7.71 x 10^-6 kg