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3.795 g of sulphur is dissolved in 100g ...

3.795 g of sulphur is dissolved in 100g of `CS_(2)`. This solution boils at 319.81 K. What is the molecular formula of sulphur in solution? The boiling point of `CS_(2)` is 319.45 kJ
(Given that `K_(b)` for `CS_(2) = 2.42 K kg mol^(-1)` and atomic mass of S = 32)

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`Delta T_(b) = (1000 k_(b) W_(2))/(W_(1)M_(2))`
or , `M_(2) = (1000 K_(b) W_(2))/(W_(1) Delta T_(b))`
`= (1000 xx 2.42 xx 3.795)/(100 xx 0.36)`
`= (91.83)/(0.36)`
`= 255.08 g`
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