1.

On dissolving 3.24 g of sulphur in 40 g of benzene, boiling point ofsolution was higher than that of benzene by 0.81K (K_(b) = 2.53 K kg mol^(-1)). What is molecular formula of sulphur ? (Atomic mass s = 32 gmol^(-1))

Answer»

SOLUTION :`M_(B) = (K_(b) xx W_(b) xx 1000)/(Delta T_(b) xx W_(A)) = (2.53 xx 3.24 xx 10^(3))/(0.81 xx 40)`
`= 253 g mol^(-1)`
Let the molecular formula of sulphur = `S_(x)`
Atomic MASS of sulphur = 32
Molecular mass = `32 xx x`
32 x = 253
`x = 7.91 = 8`
Molecular formula of sulphur = `S_(S)`


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