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The boiling point of Benzene (C(6) H(6))...

The boiling point of Benzene `(C_(6) H_(6))` is 353.23 K. When 1 . 80 g of a non - volatile solute was dissolved in 90 g of `C_(6) H_(6)` the boiling point is raised to 354 . 11 K. Calculate the molar mass of solute .
(Given `K_(b)` for benzene is 2 . 53 K kg `mol^(-1)` )

Text Solution

Verified by Experts

The correct Answer is:
`57.5 g mol^(-1)`
`Delta T_(b) = T_(b) - T_(b)^(0)`
= [354.11 - 353 .23 ] K = 0.88 K
`K_(b) = 2.53 K kg mol^(-1)`
`W_(B) = 1.80 g`
`W_(A) = 90 g = 0.09 kg`
`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 (0.09 kg))`
`= (253 xx 180)/( 88 xx 9) g mol^(-1)`
`= 57 . 5 g mol^(-1)`
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