15. A mixture of NO2(g) and NO̟(g) at 63 °C and 750 mm Hg pressure has a density of 1.98 g/L. What is the weight% NO, in the mixture? 2 There are several ways to solve this problem. In this case, the shortest way is to assume that the mole fraction of NO, = X, the mole fraction of N2O4 = 1-X. 2 Then the average Molar Mass of the mixture = Xx46+ (1-X)x92 Use this in the equation for d to obtain a linear equation in X (remember that T = 336 K): 1.98 = (750/760)x[Xx46 + (1−X)x92]/(0.08206x336) = 55.3 Xx46+ (1-X)x92 55.3 92 46 X solve for X: 101.3X = 92 →> X = 0.91 = the mole fraction of NO2 and the mole fraction of N2O4 = 1 − 0.91 = 0.09 2 The final step is to convert this to wt %: out of 1 mole mixture we have 0.91x46 g = NO2 and 0.09x92 g N₂O, total 50.1 g mixture, and the wt% of NO₂ = (0.91x46/50.1)x100% = 84 wt% NO2 and 16 wt% N₂O 2 2 2 2 4 A different way to solve the problem would be to say the partial pressure of NO is X atm, then NO (750/760-X), then the density d = [P NO2x42 +PN x92]/RT, again solve for X and from there find the mole fractions. All basically the same method. Yes, this was a challenging problem! N204 Yet another way is as in the hint: with d = 1.98 g/L we know that 1 L weighs 1.98 g. Now we can express the number of moles in that 1 L in terms of the weight % X of NO2 and 100 –X of NO: Wt. NO2 in 1 L = 0.01Xx1.98 Wt. N₂O4 in 1 L = 0.01(1-x)x1.98 Moles NO2 = 0.01Xx1.98/46 4 Moles N2O = 0.01(1-X)x1.98/92 Moles total (still in 1 L) = 0.01x1.98x[X/46 + (1-X)/92] then substitute this as n in pV=nRT for V - 1L: = (750/760)x1 = 0.01x1.98x[X/46 + (1-X)/92] x 0.08206x336 And solve for X! The answer of course will be the same.
15. A mixture of NO2(g) and NO̟(g) at 63 °C and 750 mm Hg pressure has a density of 1.98 g/L. What is the weight% NO, in the mixture? 2 There are several ways to solve this problem. In this case, the shortest way is to assume that the mole fraction of NO, = X, the mole fraction of N2O4 = 1-X. 2 Then the average Molar Mass of the mixture = Xx46+ (1-X)x92 Use this in the equation for d to obtain a linear equation in X (remember that T = 336 K): 1.98 = (750/760)x[Xx46 + (1−X)x92]/(0.08206x336) = 55.3 Xx46+ (1-X)x92 55.3 92 46 X solve for X: 101.3X = 92 →> X = 0.91 = the mole fraction of NO2 and the mole fraction of N2O4 = 1 − 0.91 = 0.09 2 The final step is to convert this to wt %: out of 1 mole mixture we have 0.91x46 g = NO2 and 0.09x92 g N₂O, total 50.1 g mixture, and the wt% of NO₂ = (0.91x46/50.1)x100% = 84 wt% NO2 and 16 wt% N₂O 2 2 2 2 4 A different way to solve the problem would be to say the partial pressure of NO is X atm, then NO (750/760-X), then the density d = [P NO2x42 +PN x92]/RT, again solve for X and from there find the mole fractions. All basically the same method. Yes, this was a challenging problem! N204 Yet another way is as in the hint: with d = 1.98 g/L we know that 1 L weighs 1.98 g. Now we can express the number of moles in that 1 L in terms of the weight % X of NO2 and 100 –X of NO: Wt. NO2 in 1 L = 0.01Xx1.98 Wt. N₂O4 in 1 L = 0.01(1-x)x1.98 Moles NO2 = 0.01Xx1.98/46 4 Moles N2O = 0.01(1-X)x1.98/92 Moles total (still in 1 L) = 0.01x1.98x[X/46 + (1-X)/92] then substitute this as n in pV=nRT for V - 1L: = (750/760)x1 = 0.01x1.98x[X/46 + (1-X)/92] x 0.08206x336 And solve for X! The answer of course will be the same.
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter12: Solutions
Section: Chapter Questions
Problem 12.92QP
Related questions
Question
Please explain this to me. I do not want just the written solution eith the method to do it. What i need is explanation to understand as the methods used are very unclear. I need help and understanding is very important. Do not just copy the methods and please explain.
![15. A mixture of NO2(g) and NO̟(g) at 63 °C and 750 mm Hg pressure has a density of 1.98 g/L. What is
the weight% NO, in the mixture?
2
There are several ways to solve this problem. In this case, the shortest way is to assume that the mole fraction of
NO, = X, the mole fraction of N2O4
= 1-X.
2
Then the average Molar Mass of the mixture = Xx46+ (1-X)x92
Use this in the equation for d to obtain a linear equation in X (remember that T = 336 K):
1.98 = (750/760)x[Xx46 + (1−X)x92]/(0.08206x336)
=
55.3 Xx46+ (1-X)x92
55.3 92 46 X
solve for X:
101.3X = 92
→>
X = 0.91 = the mole fraction of NO2 and the mole fraction of N2O4 = 1 − 0.91 = 0.09
2
The final step is to convert this to wt %: out of 1 mole mixture we have 0.91x46 g = NO2 and 0.09x92 g N₂O, total
50.1 g mixture, and the wt% of NO₂ = (0.91x46/50.1)x100% = 84 wt% NO2 and 16 wt% N₂O
2
2
2
2 4
A different way to solve the problem would be to say the partial pressure of NO is X atm, then NO
(750/760-X), then the density d = [P NO2x42 +PN x92]/RT, again solve for X and from there find the
mole fractions. All basically the same method. Yes, this was a challenging problem!
N204
Yet another way is as in the hint: with d = 1.98 g/L we know that 1 L weighs 1.98 g. Now we can
express the number of moles in that 1 L in terms of the weight % X of NO2 and 100 –X of NO:
Wt. NO2 in 1 L = 0.01Xx1.98
Wt. N₂O4
in 1 L = 0.01(1-x)x1.98
Moles NO2
= 0.01Xx1.98/46
4
Moles N2O = 0.01(1-X)x1.98/92
Moles total (still in 1 L) = 0.01x1.98x[X/46 + (1-X)/92] then substitute this as n in pV=nRT for V -
1L:
=
(750/760)x1 = 0.01x1.98x[X/46 + (1-X)/92] x 0.08206x336
And solve for X! The answer of course will be the same.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff1b8624c-c119-48ab-9e11-50a99a8168d2%2F2b0a0d0b-b017-49d7-ae3f-defe6ff2b558%2F4n784w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:15. A mixture of NO2(g) and NO̟(g) at 63 °C and 750 mm Hg pressure has a density of 1.98 g/L. What is
the weight% NO, in the mixture?
2
There are several ways to solve this problem. In this case, the shortest way is to assume that the mole fraction of
NO, = X, the mole fraction of N2O4
= 1-X.
2
Then the average Molar Mass of the mixture = Xx46+ (1-X)x92
Use this in the equation for d to obtain a linear equation in X (remember that T = 336 K):
1.98 = (750/760)x[Xx46 + (1−X)x92]/(0.08206x336)
=
55.3 Xx46+ (1-X)x92
55.3 92 46 X
solve for X:
101.3X = 92
→>
X = 0.91 = the mole fraction of NO2 and the mole fraction of N2O4 = 1 − 0.91 = 0.09
2
The final step is to convert this to wt %: out of 1 mole mixture we have 0.91x46 g = NO2 and 0.09x92 g N₂O, total
50.1 g mixture, and the wt% of NO₂ = (0.91x46/50.1)x100% = 84 wt% NO2 and 16 wt% N₂O
2
2
2
2 4
A different way to solve the problem would be to say the partial pressure of NO is X atm, then NO
(750/760-X), then the density d = [P NO2x42 +PN x92]/RT, again solve for X and from there find the
mole fractions. All basically the same method. Yes, this was a challenging problem!
N204
Yet another way is as in the hint: with d = 1.98 g/L we know that 1 L weighs 1.98 g. Now we can
express the number of moles in that 1 L in terms of the weight % X of NO2 and 100 –X of NO:
Wt. NO2 in 1 L = 0.01Xx1.98
Wt. N₂O4
in 1 L = 0.01(1-x)x1.98
Moles NO2
= 0.01Xx1.98/46
4
Moles N2O = 0.01(1-X)x1.98/92
Moles total (still in 1 L) = 0.01x1.98x[X/46 + (1-X)/92] then substitute this as n in pV=nRT for V -
1L:
=
(750/760)x1 = 0.01x1.98x[X/46 + (1-X)/92] x 0.08206x336
And solve for X! The answer of course will be the same.
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