A solution containing 1.8 g of a compound (empirical formula `CH_(2)O`) in 40 g of water is observed to freeze at `-0.465^(@)`C. The molecules formula of the compound is (`K_(f)` of water =1.86kg K`mol^(-1)`):

Options

A. `C_(2)H_(4)O_(2)`
B. `C_(3)H_(6)`
C. `C_(4)H_(8)O_(4)`
D. `C_(6)H_(12)O_(6)`

Answer

D

Explanation

`M_(2)=(1000K_(f)w_(2))/(w_(1)xxDeltaT_(f))=(1000xx1.86xx1.8)/(40xx0.465)=180`
`"E.F. mass "CH_(2)O=30 therefore n=(180)/(30)=6`
`therefore" Molecular formula "=6xxCH_(2)O=C_(6)H_(12)O_(6)`