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The molal freezing point depression cons...

The molal freezing point depression constant for benzene is `4.9 K kg mol^(-1)`. Selenium exists as a polymer of the type `Se_(n)`. When `3.26 g` selenium is dissolved in `226 g` of benzene, the observed freezing. If molecular formula of sulphur is `S_(n)`. Then find the value of `n`. (at. wt. of `S = 32`).

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The molal freezing point depression constant of benzene (C_(6)H_(6)) is 4.90 K kg mol^(-1) . Selenium exists as a polymer of the type Se_(x) . When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112^(@)C lower than pure benzene. Deduce the molecular formula of selenium. (Atomic mass of Se=78.8 g mol^(-1) )

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3.24 g of sulphur dissolved in 40g benzene, boiling point of the solution was higher than that of benzene by 0.081 K . K_(b) for benzene is 2.53 K kg mol^(-1) . If molecular formula of sulphur is S_(n) . Then find the value of n . (at.wt.of S =32 ).

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