# Question #14263

##### 1 Answer

#### 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

#color(blue)(ul(color(black)(c_"gas" = H^(cp) * P_"gas")))#

Here

#c_"gas"# is the concentration of the gasin solution#H^(cp)# isHenry's Law constant#P_"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

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

#color(blue)(ul(color(black)(PV = nRT)))#

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 theuniversal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is theabsolute temperatureof the gas

Rearrange to solve for the number of moles of gas

#PV = nRT implies n = (PV)/(RT)#

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

#n = (1.0 color(red)(cancel(color(black)("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"))))#

#n = "0.00181 moles Ar"#

You know that you dissolve this sample of argon in

#c_"gas" = "0.00181 mol L"^(-1)#

Rearrange the equation for Henry's Law to solve for

#c_"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

#H^(cp) = 1.4 * 10^(-3)color(white)(.)"mol L"^(-1)"atm"^(-1)#

so this is a fairly decent result.