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

Answer

D

Explanation

`DeltaT_(f)=K_(f)xxm`
`0.465=1.86xx(1.8)/(M)xx(1000)/(40)impliesM=180`
Molecular formula = `("empirical formula")_(n)`
`n=("Molecular mass")/("Empirical formula mass")=(180)/(30)=6`
Molecular formula = `(CH_(2)O)_(6)=C_(6)H_(12)O_(6)`.