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Raoult's law is stated as<span> the partial </span>vapor pressure<span> of each component in the </span> mixture<span> of ideal liquids is equal to the vapor pressure of the pure component multiplied by its </span>mole fraction<span> in the mixture. Mathematically, it is expressed as
P = Xi Pi
where P= vapor pressure of solution
Xi = mole fraction of i component
Pi = vapor pressure of pure i component.
In present case, P1 = 0.459 atm and n1 = number of mole of ethanol = 0.090
n2 = number of mole of </span><span> naphthalene = 0.01
</span>∴ Mole fraction of ethanol =
= 
= 0.9
Thus, vapor pressure of solution = 0.9 X 0.459 = 0.4131 atm.
Answer: <span>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.</span>
P = Xi Pi
where P= vapor pressure of solution
Xi = mole fraction of i component
Pi = vapor pressure of pure i component.
In present case, P1 = 0.459 atm and n1 = number of mole of ethanol = 0.090
n2 = number of mole of </span><span> naphthalene = 0.01
</span>∴ Mole fraction of ethanol =
Thus, vapor pressure of solution = 0.9 X 0.459 = 0.4131 atm.
Answer: <span>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.</span>