Question

The Henry’s law constant for \({\bf{O}}_2\)is \({\bf{1}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{ M at}}{{\bf{m}}^{{\bf{ - 1}}}}\)at 25 °C. What mass of oxygen would be dissolved in a 40 L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of \({\bf{O}}_2\)is 0.21 atm?

Short answer

The mass of oxygen dissolved in a 40-L aquarium at 25 °C is 0.35g. \(\)

Step-by-step solution

Definition of 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 governing equation is:

Cg = kPg

C_g= Concentration of a dissolved gas

k = Henry's Law Constant

P_g= Partial pressure of the gas

Cg = kPg

C_g = 0.0013Matm^-1x 0.21atm =2.7 x 10^-4molL^-1

The total amount =2.7 x 10^-4molL x 40L =0.0108mol

The mass of oxygen =0.0108mol x 32gmol^-1=0.346g

Mass of Oxygen with two Significant Figures = 0.35g