The vapour pressure of solution of 5 g of acetic acid in 100 g of water at `25^(@)C` was 23.40 torr and of the same mass of acetic acid in 100 g of benzene war 70 torr. Assuming acetic acid to be non-volatile, find out its physical state in the two solution, Given `P_(H_2)^(@)o=23.756 and P_(C_6H_(6))^(@)=72.5 "torr at"25^(@)C`.

Answer

Monomer in water and Dimer in benzene

Explanation

We have to calculate the observed molar mass of acetic in both solutions.
(a) in water solution.
`(P_(H_(2)O)^(@)-P_(S))/P_(S)=(eta_(CH_(3)COOH))/(eta_(H_(2)O))`
`P_(H_(2)O)^(@)=23.756" torr", P_(S)=23" torr", eta_(CH_(3)COOH)=(5g)//m_(CH_(3)COOH),`
`eta_(H_(2)O)=((100g))/((18"g mol"^(-1)))=100/18mol`
`((23.756-23.4)"torr")/(23.4"torr")=((5g))/(M_(B))xx((18"mol"^(-1)))/(100),M_(B)=(23xx(5g)xx(18"mol"^(-1)))/(0.356xx100)=59.16" g mol"^(-1)`.
(b) In banzene solution.
`P_(C_(6)H_(6))^(@)=72.5" torr", P_(S)=70"torr", n_(CH_(3)COOH)=(5g)//m_(CH_(3)COOH),`
`n_(C_(6)H_(6))=((100g))/((78"g mol"^(-1)))=((100)/(78)mol),((72.5-70)"torr")/(70"torr")=((5g))/M_(B)xx(78"mol"^(-1))/100`
`M_(B)=(70.0)/(2.5)xx(5xx78("g mol"^(-1)))/100=109" g mol"^(-1)`
(c) Predicting physical state of acetic acid in the two solutions.
`"Normal molar mass of "CH_(3)COOH=60" g mol"^(-1)`
`"Observed molar mass in water "=59.16" g mol"^(-1)`
`"Observed molar mass in benzene "= 109.2" g mol"^(-1)`
The value clearly show that acetic acid remains as monomer in water and dimer in benzene because of intermolecular hydrogen bonding.