# A mixture of CH3OH and C2H5OH acts according to the Rauolts law.In 335K, vapour pressure of pure methanol is 8.1*10^4 Pa and vapour pressure of pure ethanol is 4.5*10^4 Pa. Find the volume in...

A mixture of CH3OH and C2H5OH acts according to the Rauolts law.In 335K, vapour pressure of pure methanol is 8.1*10^4 Pa and vapour pressure of pure ethanol is 4.5*10^4 Pa. Find the volume in vapour zone in 335K for a mixture that has 64g of methanol and 46g of ethanol.

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### 1 Answer

Raoult's law can be expressed as:

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

where:

`p_(a)` = partial pressure of the component a in the mixture (or in the solution)

`p_(a) ^(o)` = vapor pressure of a pure component a

`x_(a)` = mole fraction of the component a in the mixture

The total pressure now is the sum of partial pressures of each component present in the system.

Relation this in our problem,

`p_(t otal) = p_(CH_3OH) + p_(C_2H_5OH)`

OR

`p_(t otal) = p_(CH_3OH)^(o) * (x_(CH_3OH)) + p_(C_2H_5OH)^(o) * (x_(C_2H_5OH)) `

First, we have to get the mole fractions of each component in the mixture.

`x_(CH_3OH) = (mol eCH_3OH)/(mol e CH_3OH + mole C_2H_5OH) `

`x_(CH_3OH) = (64/32)/(64/32 + 46/46) `

`x_(CH_3OH) = 2/(2+1) `

`x_(CH_3OH) = 2/3 `

`x_(C2H_5OH) = 1 -(x_(CH_3OH)) `

`x_(C2H_5OH) = 1-2/3 `

`x_(C2H_5OH) = 1/3 `

`p_(CH_3OH) =p_(CH_3OH)^(o) * (x_(CH_3OH)) `

`p_(CH_3OH) =8.1*10^4 Pa* (2/3) `

`p_(CH_3OH) = 5.4*10^4 Pa `

`p_(C_2H_5OH) =p_(C_2H_5OH)^(o) * (x_(C_2H_5OH)) `

`p_(C_2H_5OH) = 4.5*10^4 Pa *(1/3) `

`p_(C_2H_5OH) =1.5*10^4 Pa `

`p_(t otal) =5.4*10^4 Pa +1.5*10^4 Pa `

`p_(t otal) = 6.9x10^4 Pa `

To get the volume, we can use the ideal gas law, PV = nRT

`PV = nRT`

Where:

`P =6.9x10^4 Pa *(1 atm)/(1.01325x10^5 Pa) `

`= 0.681 atm`

`n =(64/32 + 46/46) `

`= 3 mol es`

R = 0.08206 atm-L/mol-K

T = 335 K

`PV = nRT`

`V = (nRT)/(P)`

`V = (3*0.08206*335)/(0.681)`

**V = 121 L answer**

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Thanks..but the answer given in my book is 78.3% CH3OH and 21.7% C2H5OH.. :confused: