# 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...

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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Asked on by shihan

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jerichorayel | College Teacher | (Level 2) Senior Educator

Posted on

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

Sources:

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