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Concentration of C02 (in mole fraction) in fat when partial pressure of C02 is 55 kPa at 25° C, is (Henry’s law constant of C02 = 8.6 x 104 torr)
- a)1.6 x 10-4
- b)4.8 x 10-3
- c)8.6 x 10-4
- d)2.4 x 10-3
Correct answer is option 'B'. Can you explain this answer?
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Concentration of C02 (in mole fraction) in fat when partial pressure o...
Note: Kh (Henry’s law constant) is in pressure unit, hence we use relation, Concentration xKH = Pressure
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Concentration of C02 (in mole fraction) in fat when partial pressure o...
Given:
Partial pressure of CO2 (P) = 55 kPa
Temperature (T) = 25°C = 298 K
Henry's law constant for CO2 (kH) = 8.6 x 104 torr
To find: Mole fraction of CO2 in fat
Solution:
Henry's law states that the amount of gas dissolved in a liquid is proportional to the partial pressure of that gas above the liquid. Mathematically,
C = kH * P
where C is the concentration of the gas in the liquid, kH is the Henry's law constant, and P is the partial pressure of the gas above the liquid.
We can convert the given partial pressure of CO2 from kPa to torr, since the Henry's law constant is given in torr.
1 kPa = 7.5 torr (approx)
Therefore,
P = 55 kPa = 55 x 7.5 torr = 412.5 torr
Now we can use Henry's law to find the concentration of CO2 in the fat.
C = kH * P
= 8.6 x 104 torr * 412.5 torr
= 3.54625 x 107 torr2
We need to convert the concentration from torr2 to mole fraction.
1 mole fraction = (1 atm / RT) * 6.022 x 1023 molecules/mole
where R is the gas constant (0.0821 L atm/mol K) and T is the temperature in Kelvin.
1 atm = 760 torr (approx)
Therefore,
1 mole fraction = (760 torr / (0.0821 L atm/mol K * 298 K)) * 6.022 x 1023 molecules/mole
= 2.419 x 10-2
Finally, we can find the mole fraction of CO2 in the fat by dividing the concentration by the total moles of all gases in the fat. Assuming that the fat is at atmospheric pressure, the total pressure is 1 atm.
Mole fraction of CO2 = C / (1 atm * 1 mole fraction)
= 3.54625 x 107 torr2 / (760 torr/mol * 2.419 x 10-2)
= 4.8 x 10-3
Therefore, the correct answer is option (B) 4.8 x 10-3.
Partial pressure of CO2 (P) = 55 kPa
Temperature (T) = 25°C = 298 K
Henry's law constant for CO2 (kH) = 8.6 x 104 torr
To find: Mole fraction of CO2 in fat
Solution:
Henry's law states that the amount of gas dissolved in a liquid is proportional to the partial pressure of that gas above the liquid. Mathematically,
C = kH * P
where C is the concentration of the gas in the liquid, kH is the Henry's law constant, and P is the partial pressure of the gas above the liquid.
We can convert the given partial pressure of CO2 from kPa to torr, since the Henry's law constant is given in torr.
1 kPa = 7.5 torr (approx)
Therefore,
P = 55 kPa = 55 x 7.5 torr = 412.5 torr
Now we can use Henry's law to find the concentration of CO2 in the fat.
C = kH * P
= 8.6 x 104 torr * 412.5 torr
= 3.54625 x 107 torr2
We need to convert the concentration from torr2 to mole fraction.
1 mole fraction = (1 atm / RT) * 6.022 x 1023 molecules/mole
where R is the gas constant (0.0821 L atm/mol K) and T is the temperature in Kelvin.
1 atm = 760 torr (approx)
Therefore,
1 mole fraction = (760 torr / (0.0821 L atm/mol K * 298 K)) * 6.022 x 1023 molecules/mole
= 2.419 x 10-2
Finally, we can find the mole fraction of CO2 in the fat by dividing the concentration by the total moles of all gases in the fat. Assuming that the fat is at atmospheric pressure, the total pressure is 1 atm.
Mole fraction of CO2 = C / (1 atm * 1 mole fraction)
= 3.54625 x 107 torr2 / (760 torr/mol * 2.419 x 10-2)
= 4.8 x 10-3
Therefore, the correct answer is option (B) 4.8 x 10-3.
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Community Answer
Concentration of C02 (in mole fraction) in fat when partial pressure o...
By Henry law p=KhX. as we find X = 55*10^3pa. /8.6*10^4* 153.32. AFTER Solving we get. b)
4.8 x 10-3
4.8 x 10-3
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Concentration of C02 (in mole fraction) in fat when partial pressure of C02 is 55 kPa at 25° C, is (Henry’s law constant of C02 = 8.6 x 104 torr)a)1.6 x 10-4b)4.8 x 10-3c)8.6 x 10-4d)2.4 x 10-3Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Concentration of C02 (in mole fraction) in fat when partial pressure of C02 is 55 kPa at 25° C, is (Henry’s law constant of C02 = 8.6 x 104 torr)a)1.6 x 10-4b)4.8 x 10-3c)8.6 x 10-4d)2.4 x 10-3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Concentration of C02 (in mole fraction) in fat when partial pressure of C02 is 55 kPa at 25° C, is (Henry’s law constant of C02 = 8.6 x 104 torr)a)1.6 x 10-4b)4.8 x 10-3c)8.6 x 10-4d)2.4 x 10-3Correct answer is option 'B'. Can you explain this answer?.
Concentration of C02 (in mole fraction) in fat when partial pressure of C02 is 55 kPa at 25° C, is (Henry’s law constant of C02 = 8.6 x 104 torr)a)1.6 x 10-4b)4.8 x 10-3c)8.6 x 10-4d)2.4 x 10-3Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Concentration of C02 (in mole fraction) in fat when partial pressure of C02 is 55 kPa at 25° C, is (Henry’s law constant of C02 = 8.6 x 104 torr)a)1.6 x 10-4b)4.8 x 10-3c)8.6 x 10-4d)2.4 x 10-3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Concentration of C02 (in mole fraction) in fat when partial pressure of C02 is 55 kPa at 25° C, is (Henry’s law constant of C02 = 8.6 x 104 torr)a)1.6 x 10-4b)4.8 x 10-3c)8.6 x 10-4d)2.4 x 10-3Correct answer is option 'B'. Can you explain this answer?.
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