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.

The vapour pressure of solution of 5 g of acetic acid in 100 g of wate

1.	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» Solution :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.