Question

learn by doing
exposing a 100.0 ml sample of water at 0°c to an atmosphere containing a gaseous solute at 20.26 kpa resulted in the dissolution of 1.45×10⁻³ g of the solute. use henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kpa.
first calculate the value for k given the mass of solute that dissolves at 20.26 kpa.
k = g kpa⁻¹
now calculate the solubility at 101.3 kpa in g/l.
c₀ = g/l
did i get this
at 0°c and 1.00 atm, as much as 0.70 g of o₂ can dissolve in 1 l of water. at 0°c and 4.00 atm, how many grams of o₂ dissolve in 1 l of water?
g

Explanation:

Step1: Calculate the value of \(k\)

Henry's law is \(C = kP\), where \(C\) is the concentration of the dissolved gas, \(k\) is Henry's law constant, and \(P\) is the partial pressure of the gas.
First, find the concentration \(C\) for the initial condition. The volume of water \(V=100.0\space mL = 0.1000\space L\), and the mass of solute \(m = 1.45\times10^{-3}\space g\). So \(C=\frac{m}{V}=\frac{1.45\times 10^{-3}\space g}{0.1000\space L}=1.45\times 10^{-2}\space g/L\)
Given \(P = 20.26\space kPa\), from \(C = kP\), we can solve for \(k\): \(k=\frac{C}{P}\)
Substitute \(C = 1.45\times 10^{-2}\space g/L\) and \(P = 20.26\space kPa\) into the formula: \(k=\frac{1.45\times 10^{-2}\space g/L}{20.26\space kPa}\approx7.16\times 10^{-4}\space g\cdot kPa^{-1}\cdot L^{-1}\)

Step2: Calculate the solubility at \(P = 101.3\space kPa\)

Using \(C = kP\) again, with \(k = 7.16\times 10^{-4}\space g\cdot kPa^{-1}\cdot L^{-1}\) and \(P = 101.3\space kPa\)
\(C=(7.16\times 10^{-4}\space g\cdot kPa^{-1}\cdot L^{-1})\times101.3\space kPa\approx0.0725\space g/L\)

Step3: Solve the "Did I Get This" problem

For the "Did I Get This" problem, using \(C = kP\). First, find \(k\) from the initial condition: when \(P_1 = 1.00\space atm\) and \(C_1=0.70\space g/L\), \(k=\frac{C_1}{P_1}=\frac{0.70\space g/L}{1.00\space atm}= 0.70\space g\cdot atm^{-1}\cdot L^{-1}\)
When \(P_2 = 4.00\space atm\), using \(C = kP\), \(C_2=(0.70\space g\cdot atm^{-1}\cdot L^{-1})\times4.00\space atm = 2.8\space g/L\)

Answer:

\(k = 7.16\times 10^{-4}\space g\cdot kPa^{-1}\), \(C_{g}=0.0725\space g/L\), and for the "Did I Get This" problem, the answer is \(2.8\space g\)