Three gaseous mixtures of N2(blue), Cl2(green), and Ne (purple) are depicted below.Which has the smallest mole fraction of N2?Which have the same mole fraction of Ne?Rank all three in order of increasing mole fraction of Cl2.

Q13-126CP - Chemistry: Molecular Nature Of Matter And Change

Q13-126CP

Question

Three gaseous mixtures of N₂(blue), Cl₂(green), and Ne (purple) are depicted below.

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

Step-by-Step Solution

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Answer

Solution C has the smallest mole fraction of N₂.
Solution A and solution C have the same mole fractions of Ne.
the order of increasing mole fraction of Cl₂is as $χ_{B} < χ_{A} < χ_{C} .$

1Step 1: Mole fraction

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;

$m o l e f r a c t i o n = \frac{(m o l e s o f o n e c o m p o n e n t)}{(t o t a l m o l e s)}$

2Step 2: Smallest mole fraction of N 2

(a)

For solution A:

Moles of nitrogen = 3mol

Total moles present in solution $= 3 m o l N_{2} + 3 m o l C l_{2} + 2 m o l N e = 8 m o l$

Mole fraction of N₂can be calculated as

$X_{A} = \frac{m o l e s o f n i t r o g e n}{t o t a l m o l e s} = \frac{3 m o l}{8 m o l} = 0.375$

For solution B:

Moles of nitrogen =4mol

Total moles present in solution $= 4 m o l N_{2} + 2 m o l C l_{2} + 4 m o l N e = 10 m o l e s$

Mole fraction of N₂can be calculated as

$X_{B} = \frac{m o l e s o f n i t r o g e n}{t o t a l m o l e s} = \frac{4 m o l}{10 m o l} = 0.4$

For solution C:

Moles of nitrogen = 4mol

Total moles present in solution $= 4 m o l N_{2} + 5 m o l C l_{2} + 3 m o l N e = 12 m o l e s$

Mole fraction of can be calculated as

$X_{C} = \frac{m o l e s o f n i t r o g e n}{t o t a l m o l e s} = \frac{4 m o l}{12 m o l} = 0.333$

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

3Step 2: Smallest mole fraction of Ne

(b)

For solution A:

Moles of Ne =2mol

Total moles present in solution $= 3 m o l N_{2} + 3 m o l C l_{2} + 2 m o l N e = 8 m o l$

Mole fraction of Ne can be calculated as

$X_{A} = \frac{m o l e s o f N_{e}}{t o t a l m o l e s} = \frac{2 m o l}{8 m o l} = 0.25$

For solution B:

Moles of Ne = 4mol

Total moles present in solution $= 4 m o l N_{2} + 2 m o l C l_{2} + 4 m o l N e = 10 m o l e s$

Mole fraction of Ne can be calculated as

$X_{B} = \frac{m o l e s o f N_{e}}{t o t a l m o l e s} = \frac{4 m o l}{10 m o l} = 0.4$

For solution C:

Moles of Ne=3mol

Total moles present in solution $= 4 m o l N_{2} + 5 m o l C l_{2} + 3 m o l N e = 12 m o l e s$

Mole fraction of Ne can be calculated as

$X_{C} = \frac{m o l e s o f N_{e}}{t o t a l m o l e s} = \frac{3 m o l}{12 m o l} = 0.25$

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

4Step 3: order of increasing mole fraction of Cl 2

(c)

For solution A:

Moles of Cl₂=3mol

Total moles present in solution $= 3 m o l N_{2} + 3 m o l C l_{2} + 2 m o l N e = 8 m o l$

Mole fraction of Cl₂can be calculated as

$X_{A} = \frac{m o l e s o f C l_{2}}{t o t a l m o l e s} = \frac{3 m o l}{8 m o l} = 0.375$

For solution B:

Moles of Cl₂=2mol

Total moles present in solution $= 4 m o l N_{2} + 2 m o l C l_{2} + 4 m o l N e = 10 m o l e s$

Mole fraction of Cl₂can be calculated as

$X_{B} = \frac{m o l e s o f C l_{2}}{t o t a l m o l e s} = \frac{2 m o l}{10 m o l} = 0.2$

For solution C:

Moles of Cl₂=5mol

Total moles present in solution $= 4 m o l N_{2} + 5 m o l C l_{2} + 3 m o l N e = 12 m o l e s$

Mole fraction of Cl₂can be calculated as

$X_{C} = \frac{m o l e s o f C l_{2}}{t o t a l m o l e s} = \frac{5 m o l}{12 m o l} = 0.42$

From the above calculation the order of increasing mole fraction of Cl₂is as $χ_{B} < χ_{A} < χ_{C}$ .

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