# Question #b629a

##### 1 Answer

#### Explanation:

For starters, it's worth mentioning that **NTP conditions** are defined as a pressure of

#color(blue)(ul(color(black)(20^@"C"))) = 20^@"C" + 273.15 = color(blue)(ul(color(black)("293.15 K")))#

Now, use the **ideal gas law equation** to calculate the **molar volume** of the gas at NTP, i.e. the volume occupied by **mole** of any ideal gas under NTP 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 equation

#V/n = (RT)/P#

Plug in your values to find

#V/n = (0.082 (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * 293.15 color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm"))))#

#V/n = "24.04 L mol"^(-1)#

This tells you that **mole** of any ideal gas occupies

You can thus say that your sample, which is said to occupy

#5.6 color(red)(cancel(color(black)("L"))) * "1 mole"/(24.04color(red)(cancel(color(black)("L")))) = "0.2329 moles"#

As you know, the **molar mass** of a substance is defined as the mass of exactly **mole** of said substance.

In your case, **mole** of this substance will have a mass of

#1 color(red)(cancel(color(black)("mole"))) * "M g"/(0.2329 color(red)(cancel(color(black)("moles")))) = (1/0.2329 * "M")# #"g"#

Therefore, you can say that the molar mass of the substance is equal to

#color(darkgreen)(ul(color(black)("molar mass" = (4.3 * "M")color(white)(.)"g mol"^(-1))))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the volume of the sample.