Avogadro's Number — Question 3

Question 3

Mass fractions must be converted before averaging molar mass

Use $N_A=6.02214076\times10^{23}\,\mathrm{mol^{-1}}$, exact. A 10.0 g gas mixture contains 40.0% helium by mass and 60.0% neon by mass. Use assigned molar masses 4.00 and 20.0 g/mol. There is no chemical reaction.

1. Find the total amount, total atom count and helium mole fraction.

2. Find mixture molar mass and compare it with the mass-weighted arithmetic average $.4(4)+.6(20)$.

Question 3 - Solution

Mass fractions must be converted before averaging molar mass

(a) Method and calculation

$$n_{\mathrm{He}}=4/4=1.00\,\mathrm{mol},\quad n_{\mathrm{Ne}}=6/20=.300\,\mathrm{mol},\quad n=1.30\,\mathrm{mol}.$$$$N=1.30N_A=7.828782988\times10^{23},\quad x_{\mathrm{He}}=1/1.3=10/13.$$

(b) Interpretation and further calculation

$$M_{\mathrm{mix}}=\frac{m}{n}=\frac{10}{1.30}=7.69231\,\mathrm{g/mol}.$$$$\frac1{M_{\mathrm{mix}}}=\frac{w_{\mathrm{He}}}{M_{\mathrm{He}}}+\frac{w_{\mathrm{Ne}}}{M_{\mathrm{Ne}}}.$$The arithmetic mass-weighted result is 13.6 g/mol and is incorrect. An arithmetic average uses mole fractions, giving $(10/13)4+(3/13)20=100/13$.

Physical check

The light species contributes a greater fraction of atoms than of mass.