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When 2.56 g of sulphur is dissolved in 1...

When 2.56 g of sulphur is dissolved in 100 g of `CS_(2)`, the freezing point of the solution gets lowerd by 0.383 K. Calculate the formula of sulphur `(S_(x))`. [Given `K_(f)` for `CS_(2)=3.83` "K kg mol"^(-1)`], [Atomic mass of sulphur=32g `mol^(-1)`]

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Calculation of molar mass of sulphur `(S_(x))`
`M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
Mass of sulphur `(W_(B))=2.56g`.
Mass of `CS_(2)(W_(A))=100g=0.1 kg`
Molal depression constant `(K_(f))=3.83 K kg mol^(-1)`
Derssion in freezing point `(DeltaT_(f))=0.383 K`
`M_(B)=((3.83 K kg mol^(-1))xx(2.56g))/((0.383K)xx(0.1 kg))=256 mol^(-1)`
Calculate of mlecular fromula of sulphr
Gram atomic mass of sulphur= `32 g mol^(-1)`
In `S_(x),xxx(32 g mol^(-1))=256 g mol^(-1)`
`x=((256 g mol^(-1)))/((32 g mol^(-1)))=8`
Molexular formula of sulphr is `S_(8)`.
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