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In some Temperature the vapour pressure of pure water is 0.0750 atm. Vapour Pressure of...

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shihan | Student, Undergraduate | Salutatorian

Posted July 7, 2013 at 4:57 PM via web

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In some Temperature the vapour pressure of pure water is 0.0750 atm. Vapour Pressure of a solution that has 5g of Organic Compound and 90g of Water is 0.0747 atm. Find the molecular weight of the organic compound.

 

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted July 7, 2013 at 5:33 PM (Answer #1)

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At a particular temperature, 

`p_(H_2O) = p_(H_2O)^o *(x_(H_2O))`

Where:

`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 

 

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