# Question #d2ce6

##### 1 Answer

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

#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**.