Question

Question #b629a

Answer 1

`"4.3M g mol"^(-1)`

Explanation:

For starters, it's worth mentioning that NTP conditions are defined as a pressure of `color(blue)(ul(color(black)("1 atm")))` and a temperature 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 `1` 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 the universal gas constant, equal to `0.0821("atm L")/("mol K")`
• `T` is the absolute temperature of 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 `1` mole of any ideal gas occupies `"24.04 L"` under NTP conditions.

You can thus say that your sample, which is said to occupy `"5.6 L"` under these conditions, will contain

`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 `1` mole of said substance.

In your case, `1` 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.