The vapour pressure of pure ethanol at 60 °C is 0.459 atm. If 10 millimoles of naphthalene (non-volatile) is dissolved in 90 millimoles of ethanol the vapour pressure will be
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Thus, vapor pressure of solution = 0.9 X 0.459 = 0.4131 atm. Answer: 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 0.4131 atm.
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