Brandon Snider Posted December 12, 2012 Posted December 12, 2012 I am faced with the question: The vapor pressure of pure ethanol at 60 °C is 0.459 atm. Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile) in 90.0 mmol ethanol will have a vapor pressure of __________ atm. If I convert the milimolar to molar I get: 0.01 mol Nap. 0.09 mol Eth. and 0.10 mol total Thus the mole fractions should be: X_Nap. = 0.01 / 0.10 = 0.10 X_ Eth. = 0.09 / 0.10 = 0.90 Which according to Raoult's Law Vapor Pressure of Soln. = Sum of Vapor Partial pressures 0.459 atm * X_Nap. -{0.10} or 0.459 * 0.10 = 0.0459 Partial Pressure Nap. 0.459 atm * X_Eth. -{0.90} or 0.459 * 0.90 = 0.4131 Partial Pressure Eth. I sum the two and get: 0.459. Which is consequently the same as the pure vapor pressure. I am assuming that this is related to the fact that naphthalene is supposed to be non-volatile. My problem is that this matches none of the answer choices. Any help appreciated. My exam is in 20 mins. Answer Choices: 0.498 0.0918 0.367 0.413 0.790 *Chem software says that 0.413 is the answer. Not sure how.
John Cuthber Posted December 12, 2012 Posted December 12, 2012 Naphthalene stinks (I know it's too late for the exam)
Tawfik Posted December 13, 2012 Posted December 13, 2012 Hi,they said in the question that: naphthalene is nonvolatile (the nonvolatile compound don't have a vapor pressure..)...so in ur exercise solution u shouldn't take consideration of naphta's pressure.....:0.459 atm * X_Nap. -{0.10} or 0.459 * 0 = 0 Partial Pressure Nap.0.459 atm * X_Eth. -{0.90} or 0.459 * 0.90 = 0.4131 Partial Pressure Eth. is the vapor pressure of solution.
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