How many molecules of ethane gas, C_2H_6C2H6, are in 33.6 L at STP?

1 Answer
Aug 17, 2016

Well, one mole of ideal gas at "STP"STP occupies 22.4*L22.4L.

Explanation:

So, assuming ethane behaves ideally, 33.6*L33.6L of the stuff represents (33.6*L)/(22.4*L*mol^-1)33.6L22.4Lmol1 == 1.5*mol1.5mol.

Now, it is a fact that 11 molmol of any substance contains "Avogadro's Number, "N_AAvogadro's Number, NA of particles, i.e. 6.022xx10^236.022×1023 particles.

And thus "number of ethane gas molecules "=6.022xx10^23*cancel(mol^-1)xx1.5*cancel(mol) ~= 9xx10^23 "ethane formula units".