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In some Temperature the vapour pressure of pure water is 0.0750 atm. Vapour Pressure of...
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At a particular temperature,
`p_(H_2O) = p_(H_2O)^o *(x_(H_2O))`
`p_(H_2O)` = partial pressure of water in the solution
`p_(H_2O)^o` = vapor pressure of pure water
`x_(H_2O)` = mole fraction of water in the solution
`p_(H_2O) = p_(H_2O)^o *(x_(H_2O)) `
`0.0747 = 0.0750 *(x_(H_2O)) `
`x_(H_2O) = 0.0747/0.0750 `
`x_(H_2O) = 0.996 `
Let us make x be equal to the molecular weight of the unknown compound.
`x_(H_2O) = (mol es H_2O)/(mol es H_2O + mol es Compound)`
`0.996 = (90/18)/(90/18 + 5/x)`
`0.996 = 5/(5 + 5/x) `
`5 + 5/x = 5/0.996 `
`5 + 5/x = 5.02008 `
`5/x = 5.02008-5 `
`5/x = 0.02008 `
`x = 5/0.02008 `
x = 249 grams/mole
Posted by jerichorayel on July 7, 2013 at 5:33 PM (Answer #1)
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