Question

Question #d2ce6

Answer 1

Here's what I got.

Explanation:

The first thing you need to do here is to calculate the partial pressure of argon by using the ideal gas law equation

`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 the universal gas constant, equal to `0.0821("atm L")/("mol K")`
• `T` is the absolute temperature of the gas

Now, in order to convert the sample to moles, use the molar mass of argon

`1.40 color(red)(cancel(color(black)("g"))) * "1 mole Ar"/(39.948color(red)(cancel(color(black)("g")))) = "0.03505 moles Ar"`

Rearrange the ideal gas law equation to solve for `P`

`PV = nRT implies P = (nRT)/V`

then plug in your values to find -- do not forget to convert the temperature to Kelvin!

`P = (0.03505 color(red)(cancel(color(black)("moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(1.00color(red)(cancel(color(black)("L"))))`

`P = color(darkgreen)(ul(color(black)("0.856 atm")))`

Now, according to Dalton's Law of Partial Pressures, the total pressure of a gaseous mixture is equal to the sum of the partial pressure of each component of said mixture.

This means that after you add the ethane vapor to the flask, the total pressure will be

`P_"total" = P_"Ar" + P_"ethane"`

You can thus say that the partial pressure of ethane will be

`P_"ethane" = "1.150 atm" - "0.856 atm" = color(darkgreen)(ul(color(black)("0.294 atm")))`

I'll leave both answers rounded to three sig figs.