Question

At 0 °C and 1.00 atm, as much as 0.70 g of \({\bf{O}}_2\)can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of \({\bf{O}}_2\)dissolve in 1 L of water?

Short answer

The amount of \({\rm{O2}}\) dissolved in 1L of water= 2.80g

Step-by-step solution

Henry’s Law

Henry’s law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

Explanation

This problem requires the application of Henry’s law.

The main equation is:

C_g = kP_g

C = Concentration of a dissolved gas

k = Henry's Law Constant

P = Partial pressure of the gas

k = C_g/P_g

k = 0.70g/1.00atm =0.70g atm^-1

Under the new conditions,

C_g=0.70g atm^-1x 4.00atm = 2.80g