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satela [25.4K]
3 years ago
9

At a given temperature the vapor pressure of pure liquid diethyl ether and isopropyl alcohol are 730 torr and 310 torr, respecti

vely. a solution prepared by mixing isopropyl alcohol and diethyl ether obeys raoult's law. at this temperature, what is the vapor pressure of diethyl ether over a solution in which its mole fraction is 0.270?
Chemistry
1 answer:
lozanna [386]3 years ago
8 0

Answer:

197.1 torr.

Explanation:

  • Raoult's law states that <em>the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture</em>.
  • <em>Pi = Xi.Pi°</em>, where, Pi is the partial vapor pressure of the component i, Xi is the mole fraction of the component i, and Pi° is the vapor pressure of the pure component i.
  • For diethyl ether; Xi = 0.270, and Pi° = 730.0 torr.
  • So, the vapor pressure of diethyl ether over the solution (Pi diethyl ether) = (0.270 x 730.0) = 197.1 torr.
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The variation in bond angles of  unsaturated aliphatic hydrocarbons  can be explained by two concepts; The valence shell electron pair repulsion (VSEPR) model and hybridization.

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Hybridization is the mixing or blending of two or more pure atomic orbitals (s,p and d) to form two or more hybrid atomic orbitals that are identical in shape and energy . e.g sp, sp² , sp³  hybrid orbitals etc .

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