When 30.0 g of a non - volatile solute having the empirical fromual `CH_(2)O` are dissolved in 800g of water , the solution freezes at ` - 1.16^(@)C` . What is the molecular formula of the solute ? (`k_(f)` for water `= 1.86 K m^(-1)`)

Answer

`C_(2)H_(4)O_(2)`

Explanation

`M_(B) = (k_(f) xx w_(B) xx 1000)/(w_(A) xx Delta T_(f)) = (1.86 xx 30.0 xx 1000)/(800 xx 1.16)`
`M_(B) = 60.1`
` n = ("Mol.formula mass")/("Empirical formual mass") = (60)/(30) = 2`
Molecular formual ` = (CH_(2)O)_(2) = C_(2)H_(4)O_(2)` .