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 formulea 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_(5)H_(10)O_(5)`

Explanation

`M_(2) = (1000 xx k_(f) xx w_(2))/(w_(1) xx Delta T_(f))`
`Delta T_(f) = 0- (-0.465) = 0.465`
`M_(2) = (1000 xx 1.86 xx 1.8)/(40 xx 0.465) = 180`
Empirical formual mass ` = 12+2xx1+16 = 30`
`n = ("Molecular mass ")/("Empirical formula mass")`
` = (180)/(30) = 6`
`therefore` Molecular formual `(CH_(2)O)_(6) = C_(6)H_(12)O_(6)`.