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)` )

Explanation

`DeltaT_b = 354.11-353.23`
= 0.88
`W_B = 1.80 g`
`W_A = 90 g`
`K_b=2.53" kg mol"^(-1)`
`DeltaT_b=(K_bxxW_Bxx1000)/(W_AxxDeltaT_b)`
`M_B=(K_bxxW_Bxx1000)/(W_AxxDeltaT_b)`
`=(2.53xx1.80xx1000)/(90xx0.88)`
`= (4554)/(90xx0.88)`
`=57.5 " g mol"^(-1)`