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If ethanol and chloroform are present in a molar ratio of 2:3 then what is the vapor pressure at 20° C if vapor pressures of pure liquids are 5.95 kPa and 21.17 kPa, respectively?
- a)16.692 kPa
- b)15.082 kPa
- c)8.731 kPa
- d)12.038 kPa
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
If ethanol and chloroform are present in a molar ratio of 2:3 then wha...
Given,
P0eth = 5.95 kPa
P0chl = 21.17 kPa
Mole ratio of ethanol ∶ chloroform = 2 ∶ 3
Total number of parts = 2 + 3 = 5
Therefore, mole fraction of ethanol, Xeth = 2/5 = 0.4
Mole fraction of chloroform, Xchl = 3/5 = 0.6
From Raoult’s law, pA = p0A x XA
Peth = 5.95 x 0.4 = 2.38 kPa
Pchl = 21.17 x 0.6 = 12.702 kPa
From Dalton’s law, Ptotal = Peth + Pchl
Ptotal = 2.38 + 12.702 = 15.082 kPa.
P0eth = 5.95 kPa
P0chl = 21.17 kPa
Mole ratio of ethanol ∶ chloroform = 2 ∶ 3
Total number of parts = 2 + 3 = 5
Therefore, mole fraction of ethanol, Xeth = 2/5 = 0.4
Mole fraction of chloroform, Xchl = 3/5 = 0.6
From Raoult’s law, pA = p0A x XA
Peth = 5.95 x 0.4 = 2.38 kPa
Pchl = 21.17 x 0.6 = 12.702 kPa
From Dalton’s law, Ptotal = Peth + Pchl
Ptotal = 2.38 + 12.702 = 15.082 kPa.
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Community Answer
If ethanol and chloroform are present in a molar ratio of 2:3 then wha...
In order to determine the vapor pressure at 20°C, we need to know the vapor pressure of each component individually and their mole fractions.
Let's assume that the vapor pressure of ethanol at 20°C is 43.9 mmHg and the vapor pressure of chloroform at 20°C is 212.8 mmHg. These values can be obtained from reference sources or experimental data.
Given that the molar ratio of ethanol to chloroform is 2:3, we can determine the mole fraction of each component:
Mole fraction of ethanol (Xethanol) = moles of ethanol / total moles
Mole fraction of ethanol (Xethanol) = 2 / (2 + 3) = 0.4
Mole fraction of chloroform (Xchloroform) = moles of chloroform / total moles
Mole fraction of chloroform (Xchloroform) = 3 / (2 + 3) = 0.6
Now, we can calculate the partial pressure of each component using the ideal gas law:
Partial pressure of ethanol (Pethanol) = Xethanol * vapor pressure of ethanol
Partial pressure of ethanol (Pethanol) = 0.4 * 43.9 mmHg = 17.56 mmHg
Partial pressure of chloroform (Pchloroform) = Xchloroform * vapor pressure of chloroform
Partial pressure of chloroform (Pchloroform) = 0.6 * 212.8 mmHg = 127.68 mmHg
Finally, we can calculate the total vapor pressure at 20°C by summing up the partial pressures:
Total vapor pressure = Pethanol + Pchloroform
Total vapor pressure = 17.56 mmHg + 127.68 mmHg = 145.24 mmHg
Therefore, the vapor pressure at 20°C for a mixture of ethanol and chloroform in a molar ratio of 2:3 is approximately 145.24 mmHg.
Let's assume that the vapor pressure of ethanol at 20°C is 43.9 mmHg and the vapor pressure of chloroform at 20°C is 212.8 mmHg. These values can be obtained from reference sources or experimental data.
Given that the molar ratio of ethanol to chloroform is 2:3, we can determine the mole fraction of each component:
Mole fraction of ethanol (Xethanol) = moles of ethanol / total moles
Mole fraction of ethanol (Xethanol) = 2 / (2 + 3) = 0.4
Mole fraction of chloroform (Xchloroform) = moles of chloroform / total moles
Mole fraction of chloroform (Xchloroform) = 3 / (2 + 3) = 0.6
Now, we can calculate the partial pressure of each component using the ideal gas law:
Partial pressure of ethanol (Pethanol) = Xethanol * vapor pressure of ethanol
Partial pressure of ethanol (Pethanol) = 0.4 * 43.9 mmHg = 17.56 mmHg
Partial pressure of chloroform (Pchloroform) = Xchloroform * vapor pressure of chloroform
Partial pressure of chloroform (Pchloroform) = 0.6 * 212.8 mmHg = 127.68 mmHg
Finally, we can calculate the total vapor pressure at 20°C by summing up the partial pressures:
Total vapor pressure = Pethanol + Pchloroform
Total vapor pressure = 17.56 mmHg + 127.68 mmHg = 145.24 mmHg
Therefore, the vapor pressure at 20°C for a mixture of ethanol and chloroform in a molar ratio of 2:3 is approximately 145.24 mmHg.
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