Question

When a gas is bubbled through water at 298 k,A very dilute solution of the gas at 298 k is 100 kbar.if the gas exerts a partial pressure of 1 bar, the number of millimoles Of the gas dissolved in 1 L of water is?

Answer 1

Hey mate, here is your answer :

Assume that N2 exerts a partial pressure of - 106… ... millimoles of N2 gas would dissolve in 1 litre of water. ... Henry's law constant for N2 at 293 K is 76.48 kbar.

Hope this answer helps you...

Answer 2

Explanation:

Relation between partial pressure, number of moles and pressure is as follows.

                           p = $x \times P$

where,        p = partial pressure

                   x = number of moles

                   P = pressure

Since, partial pressure is given as 1 bar, and pressure is 100 kbar which is equal to 100000 bar.

Therefore, calculate the number of moles using the above formula as follows.

                            p = $x \times P$

                          1 bar = $x \times 100000 bar$

                            x = $10^{-5}$

So, in 1 liter of water there are $\frac{1000 ml}{18}$ equals 55.5 moles.

                       x = $\frac{n}{55.5+n}$

               $10^{-5}$ = $\frac{n}{55.5+n}$

                            n = 0.000555 moles

1 mole equals 1000 milimoles. Hence, 0.000555 moles equal 0.555 milimoles.