How many molecules of neon are associated with a 75*L75L volume of neon gas?

1 Answer
May 16, 2016

The question should have specified a temperature. I am going to assume that the gas is at room temperature, 298*K298K.

Explanation:

We know that at room temperature and pressure, 11 molmol of ideal gas occupies a volume of 25.4*L25.4L.

We assume (reasonably) that NeNe behaves ideally.

Thus the number of moles of NeNe == (75*L)/(25.4*L*mol^-1)75L25.4Lmol1 ~= 3*mol3mol.

But by definition, 1*mol1mol of stuff possesses 6.022xx10^236.022×1023 individual items of that stuff. If you can grasp this concept, it will save you a ton of angst.

Thus "Molecules (atoms) of neon"Molecules (atoms) of neon == (75*L)/(25.4*L*mol^-1)xx6.022xx10^23" neon atoms per mole"75L25.4Lmol1×6.022×1023 neon atoms per mole

~=18xx10^23 " neon atoms"18×1023 neon atoms.

What is the mass of this quantity of neon gas?

(Neon gas is a monatomic molecule).

See also here.