InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
`1.22 g` of benzoic acid is dissolved in acetone and benzene separately. Boiling point of mixture with acetone increase by `0.17^(@)C` and boiling point of mixture with benzene increases by `0.13^(@)C`. `K_(b) ("acetone")=1.7 K kg "mol"^(-1)`, Mass of acetone `= 100 g`, `K_(b) ("benzene")=2.6 k Kg mol^(-1)`, Mass of benzene `= 100 g`, Find molecular weight of benzoic acid in acetone and in benzene solution. Justify your answer with structure. |
|
Answer» Correct Answer - `122,224` (i) In first case, `DeltaT_(b)=K_(b)xxm=K_(b)xx("Wt. of solute")/("Mol.wt. of solute")xx(1000)/(" wt. of solvent")` or `0.17=1.7xx(1.22)/(Mxx100xx10^(-3))` `M=122gm//"mole"` Thus the benzoic acid exists as a monomer in acetone (ii) In second case, `DeltaT_(b)=K_(b)xx("Wt. of solute")/("Mol. wt. of solute")xx(1000)/("wt. of solvent")` or `0.13=2.6xx(1.22)./(Mxx100xx10^(-3))` `M=224` Double molecular weight of benzoic acid `(244)` in acetone solution indicates that benzoic acid exists as dimer in acetone. |
|