Three gaseous mixtures ofN₂(blue), Cl₂(green), and Ne (purple) are depicted below.

1. Which has the smallest mole fraction of N₂?
2. Which have the same mole fraction of Ne?
3. Rank all three in order of increasing mole fraction of Cl₂.

Mole fraction is the ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all the components. It can be calculated as follow;

$mole\hspace{0.33em}fraction=\frac{(moles\hspace{0.33em}of\hspace{0.33em}one\hspace{0.33em}component)}{(total\hspace{0.33em}moles)}$

(a)

For solution A:

Moles of nitrogen = 3mol

Total moles present in solution$=3mol\hspace{0.33em}{N}_2+3mol\hspace{0.33em}C{l}_2+2mol\hspace{0.33em}Ne=8mol$

Mole fraction of N₂can be calculated as

$$\begin{gathered} X_A=\frac{molesofnitrogen}{totalmoles} \\ =\frac{3mol}{8mol} \\ =0.375 \end{gathered}$$

For solution B:

Moles of nitrogen =4mol

Total moles present in solution$=4mol\hspace{0.33em}{N}_2+2mol\hspace{0.33em}C{l}_2+4mol\hspace{0.33em}Ne=10moles$

Mole fraction of N₂can be calculated as

role="math" localid="1663401770443" $$\begin{gathered} X_B=\frac{molesofnitrogen}{totalmoles} \\ =\frac{4mol}{10mol} \\ =0.4 \end{gathered}$$

For solution C:

Moles of nitrogen = 4mol

Total moles present in solution$=4mol\hspace{0.33em}{N}_2+5mol\hspace{0.33em}C{l}_2+3mol\hspace{0.33em}Ne=12moles$

Mole fraction of can be calculated as

$$\begin{gathered} X_C=\frac{molesofnitrogen}{totalmoles} \\ =\frac{4mol}{12mol} \\ =0.333 \end{gathered}$$

Hence, from the above calculations it is clear that solution C has the smallest mole fraction of N₂.

(b)

For solution A:

Moles of Ne =2mol

Total moles present in solution $=3mol\hspace{0.33em}{N}_2+3mol\hspace{0.33em}C{l}_2+2mol\hspace{0.33em}Ne=8mol$

Mole fraction of Ne can be calculated as

$$\begin{gathered} X_A=\frac{molesof{N}_e}{totalmoles} \\ =\frac{2mol}{8mol} \\ =0.25 \end{gathered}$$

For solution B:

Moles of Ne = 4mol

Total moles present in solution$=4mol\hspace{0.33em}{N}_2+2mol\hspace{0.33em}C{l}_2+4mol\hspace{0.33em}Ne=10moles$

Mole fraction of Ne can be calculated as

localid="1663402516668" $$\begin{gathered} X_B=\frac{molesof{N}_e}{totalmoles} \\ =\frac{4mol}{10mol} \\ =0.4 \end{gathered}$$

For solution C:

Moles of Ne=3mol

Total moles present in solution$=4mol\hspace{0.33em}{N}_2+5mol\hspace{0.33em}C{l}_2+3mol\hspace{0.33em}Ne=12moles$

Mole fraction of Ne can be calculated as

$$\begin{gathered} X_C=\frac{molesof{N}_e}{totalmoles} \\ =\frac{3mol}{12mol} \\ =0.25 \end{gathered}$$

So, from the above calculation it is clear that solution A and solution C have the same mole fractions of Ne.

(c)

For solution A:

Moles of Cl₂=3mol

Total moles present in solution $=3mol\hspace{0.33em}{N}_2+3mol\hspace{0.33em}C{l}_2+2mol\hspace{0.33em}Ne=8mol$

Mole fraction of Cl₂can be calculated as

$$\begin{gathered} X_A=\frac{molesofC{l}_2}{totalmoles} \\ =\frac{3mol}{8mol} \\ =0.375 \end{gathered}$$

For solution B:

Moles of Cl₂=2mol

Total moles present in solution$=4mol\hspace{0.33em}{N}_2+2mol\hspace{0.33em}C{l}_2+4mol\hspace{0.33em}Ne=10moles$

Mole fraction of Cl₂can be calculated as

$$\begin{gathered} X_B=\frac{molesofC{l}_2}{totalmoles} \\ =\frac{2mol}{10mol} \\ =0.2 \end{gathered}$$

For solution C:

Moles of Cl₂=5mol

Total moles present in solution$=4mol\hspace{0.33em}{N}_2+5mol\hspace{0.33em}C{l}_2+3mol\hspace{0.33em}Ne=12moles$

Mole fraction of Cl₂can be calculated as

$$\begin{gathered} X_C=\frac{molesofC{l}_2}{totalmoles} \\ =\frac{5mol}{12mol} \\ =0.42 \end{gathered}$$

From the above calculation the order of increasing mole fraction of Cl₂is as ${\chi }_B<{\chi }_A<{\chi }_C$.