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
color(blue)(ul(color(black)(PV = nRT)))
Here
P is the pressure of the gasV is the volume it occupiesn is the number of moles of gas present in the sampleR is the universal gas constant, equal to0.0821("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
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
In your case,
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.