Boiling point of benzene is 353.23 K . When 1 . 80 g of non-volatile solute was dissolved in 90 g of benzene the boiling point is raised to 354. 11 K? Calculate molar mass of solute .
(`K_(b)` for benzene is 2 . 53 K kg ` mol^(-1)`)

Answer

`57.5g mol^(-1) `

Explanation

`Delta T_(b) = T_(b) - T_(b)^(@)`
`= (354 . 11) - 354 . 23 ) K`
0 . 88 K
`W_(B) = 1 . 80 g, W_(A) = 90 g = 0 . 0 9 kg`
`K_(b) = 2 . 53 K kg mol^(-1) , M_(B) = ?`
`Delta T_(b) = K_(b) xx (W_(B))/( M_(B)) xx (1)/( W_(A) (kg))`
`M_(B) = (K_(b) xx W_(B))/( Delta T_(b) xx W_(A) (kg))`
`= (( 2 . 53 K kg mol^(-1)) xx (1 . 80 g)) /( ( 0 . 88 K) xx ( 0 . 0 9 kg))`
`= (253 xx 180)/( 88 xx 9) g mol^(-1)`
`= 57 . 5 g mol^(-1)`