# Question #df940

##### 1 Answer

#### Answer:

#### Explanation:

The tricky part here is actually to get the conditions for *Room Temperature and Pressure* (RTP) right. Since you didn't provide more info about the RTP conditions given to you, I'll work with a temperature of

Now, the idea is that you can use the **ideal gas law** equation to find the **molar volume of a gas** at RTP, i.e. the volume occupied by **mole** of gas under these conditions.

#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 the ideal gas law equation

#V/n = (RT)/P#

Plug in your values -- **do not** forget to convert the temperature to *Kelvin*!

#V/n = (0.0821(color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("atm"))))#

#V/n = "24.45 L mol"^(-1)#

This tells you that under RTP conditions, **mole** of any ideal gas occupies

#color(blue)(ul(color(black)("1 dm"^3 = "1 L")))#

you can say that it contains

#1 color(red)(cancel(color(black)("L"))) * "1 mole gas"/(24.45color(red)(cancel(color(black)("L")))) = "0.04090 moles gas"#

Finally, you know that this sample has a mass of **mole** will be equal to

#1 color(red)(cancel(color(black)("mole"))) * "1.58 g"/(0.04090color(red)(cancel(color(black)("moles")))) = color(darkgreen)(ul(color(black)("38.6 g")))#

I'll leave the answer rounded to three **sig figs**.

Notice that you've essentially determined the **molar mass** of the gas, i.e. the mass of **mole** of this gas.

**SIDE NOTE** *Notice that you can get to the result a bit faster by rearranging the ideal gas law equation as*

#n/V = P/(RT)#

*This will give you the number of moles present in one liter of gas kept under RTP conditions. Once again, you will have*

#n/V = (1 color(red)(cancel(color(black)("atm"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))#

#n/V = "0.04090 mol L"^(-1)#

*Since you know that your sample occupies* *you can say that it contains* **moles** of gas.