Addition of 0.643g of a compound to 43.95g of benzene lowers the freezing point from `5.51^@`C to `5.03^@`C. If `K_f` for benzene is 5.12K kg `mol^-1` , calculate the molar mass of the compound.

Answer

`156 g mol^(-1)`

Explanation

` W_(b) = 0 . 643 g`
`W_(A) = 50 mL xx 0 . 879 g mL^(-1)`
`= 43 . 95 g = 0 . 0 44 kg`
`Delta T_(f) = ( 5 . 51 - 5 . 0 3 )^(@) C `
`= 0 . 48^(@) C = 0 . 48 K`
`k_(f) = 5 . 12 kg mol^(-1) , M_(B)` = ?
`Delta T_(f) = K_(f) xx (W_(B))/( M_(B)) xx (1)/( W_(A) (kg))`
`M_(B) = (K_(f) xx w_(B))/( Delta T_(f) xx W_(A) (kg))`
`= ( 512 xx K kg mol^(-1) xx ( 0 . 0643 g))/( ( 0 . 48 K) xx ( 0 . 0 44 kg))`
`= (512 xx 643)/( 48 xx 44) = 155 . 88 g mol^(-1)`