Question

A solution containing $1.8\, g$ of a compound (empirical formula $ C{{H}_{2}}O $ ) in $40\, g$ of water is observed to freeze at -$0.465^\circ C$. The molecular formula of the compound is ( $ {{k}_{f}} $ of water = $ 1.86\text{ }kg\,K\,mo{{l}^{-1}} $ )

• A. $ {{C}_{2}}{{H}_{4}}{{O}_{2}} $
• B. $ {{C}_{3}}{{H}_{6}}{{O}_{3}} $
• C. $ {{C}_{4}}{{H}_{8}}{{O}_{4}} $
• D. $ {{C}_{5}}{{H}_{10}}{{O}_{5}} $
• E. $ {{C}_{6}}{{H}_{12}}{{O}_{6}} $

Answer 1

Answer: (E)

$ \Delta {{T}_{f}}={{K}_{f}}\times m $
$ \because $ $ \Delta {{T}_{f}}=T_{f}^{o}-{{T}_{f}} $
$ =0{}^\circ -(-0.465{}^\circ )=0.465{}^\circ C $
$ \therefore $ $ 0.465=1.86\times \frac{1.8}{M}\times \frac{1000}{40} $
or $ M=180 $ empirical formula weight
$ =12+1\times 2+16 $
$ =30 $
$ \because $ $ n=\frac{molecular\text{ }weight}{empirical\text{ }formula\text{ }weight}=\frac{180}{30}=6 $
$ \therefore $ Molecular formula $ ={{(C{{H}_{2}}O)}_{6}} $
$ ={{C}_{2}}{{H}_{12}}{{O}_{6}} $
Thus, molecular formula of compound
$ ={{C}_{6}}{{H}_{12}}{{O}_{6}} $