Following are equimolal (`=`f equimolar) aqueous solutions
(A) `1 m glucose `
(B) `1m NaCl` (C) `1m BaCl_(2)`
(D) `1m Na_(3)PO_(4)` (E) `1m` benzoic acid.
assume `100%` ionisation in `B,C,D` and `100%` dimer formation in `E` arrange them in increasing (1) boiling point, (2) freezing point, (3) osmotic pressure, (4) vapour pressure.\

Answer

`DltCltBltAltE`

Explanation

(1) Boiling point of solution
`T=T_(0)+(DeltaT_(b))`
`DeltaT_(b)=mK_(b)i` ltbRgt Thus, greater the value of `i`, greater the value of `DeltaT_(b)` and boiling point of solution.
`therefore EltAltBltCltD`
(2) Freezing pont of solution
`T=T_(B)-(DeltaT_(f))`
`DeltaT_(f)=mK_(f)i`
Thus, greater the value of `i`, greater the value of `DeltaT_(f)` but smaller the value of freezing point.
Thus, `EltAltBltCltD`
(4) Vapour pressure of solution
`P_("solution")=P_("solvent")^(0)-DeltaP`
`(DeltaP)/(P^(0))=X_("solute")=(n_(1)i)/(n_(1)i+n_(2))=(n_(1)i)/(n_(2))`
`therefore` The greater the value of `i`, greater the value of `Delpap` hence, smaller the value of vapour pressure of solution.
Thus, `DltCltBltAltE`.