Question #b629a

1 Answer
May 22, 2017

"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.