# Question 14263

Mar 4, 2017

$1.8 \cdot {10}^{- 3} {\text{mol L"^(-1)"atm}}^{- 1}$

#### Explanation:

Henry's Law states that the solubility of a gas is directly proportional to the partial pressure of the gas above the solution.

Mathematically, this is expressed as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{c}_{\text{gas" = H^(cp) * P_"gas}}}}}$

Here

• ${c}_{\text{gas}}$ is the concentration of the gas in solution
• ${H}^{c p}$ is Henry's Law constant
• ${P}_{\text{gas}}$ is the partial pressure of the gas above the solution

Now, you already know the partial pressure of the gas above the solution because you are told that the gas was dissolved at a pressure of $\text{1 atm}$.

The first thing to do here will be to use the ideal gas law equation to figure out the number of moles of argon dissolved in solution

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Rearrange to solve for the number of moles of gas

$P V = n R T \implies n = \frac{P V}{R T}$

Plug in your values to find the value of $n$ -- do not forget to convert the temperature from degrees Celsius to Kelvin

$n = \left(1.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 4.43 * 10^(-2)color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K}}}}\right)$

$n = \text{0.00181 moles Ar}$

You know that you dissolve this sample of argon in $\text{1.0 L}$ of water, which means that the concentration of the dissolved gas will be

${c}_{\text{gas" = "0.00181 mol L}}^{- 1}$

Rearrange the equation for Henry's Law to solve for ${H}^{c p}$

${c}_{\text{gas" = H^(cp) * P_"gas" implies H^(cp) = c_"gas"/P_"gas}}$

Plug in your values to find

H^(cp) = "0.00181 mol L"^(-1)/"1.0 atm" = color(darkgreen)(ul(color(black)(1.8 * 10^(-3)color(white)(.)"mol L"^(-1)"atm"^(-1))))#

The answer is rounded to two sig figs and expressed in scientific notation.

The cited value for Henry's Law constant for Argon in water at ${25}^{\circ} \text{C}$ is

${H}^{c p} = 1.4 \cdot {10}^{- 3} \textcolor{w h i t e}{.} {\text{mol L"^(-1)"atm}}^{- 1}$

so this is a fairly decent result.