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slava [35]
3 years ago
15

The vapor pressure of pure ethanol at 60 °c is 0.459 atm. raoult's law predicts that asolution prepared by dissolving 10.0 mmol

naphthalene (nonvolatile) in 90.0 mmol ethanol willhave a vapor pressure of ________ atm.
Chemistry
1 answer:
Vlad1618 [11]3 years ago
6 0
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 = \frac{n1}{\text{n1 + n2}} = \frac{0.090}{\text{0.090 + 0.010}}  = 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>
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