At 23 oC the solubility of a gas sample is 0.05 g/L in water at 1.0 atm. What is the solubility when the pressure increases to 2.0 atm?

I don't know what formula to use for this question.

Henry's Law applies, doesn't it?

you increase pressure by two times, solubility increases twice.

http://en.wikipedia.org/wiki/Henry%27s_law

To solve this question, you can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The mathematical expression for Henry's Law is:

C = k * P

where C is the solubility of the gas in the liquid, k is the proportionality constant, and P is the pressure of the gas above the liquid.

In this case, you know that at 1.0 atm, the solubility of the gas is 0.05 g/L. Let's denote this solubility as C1 and the pressure as P1. Now you need to find the solubility at 2.0 atm, which we will denote as C2 and the pressure as P2.

Using Henry's Law, you can set up the equation:

C1 / P1 = C2 / P2

Plugging in the given values, we have:

0.05 g/L / 1.0 atm = C2 / 2.0 atm

To find C2, we can rearrange the equation:

C2 = (0.05 g/L / 1.0 atm) * 2.0 atm

Now you can calculate the solubility C2.