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On dissolving 3.24 g of sulphur in 40 g ...

On dissolving `3.24 g` of sulphur in `40 g` of benzene, the boiling point of the solution was higher than sulphur? (`K_(b)` for benzene = `2.53 K kg mol^(-1)` , atomic mass of sulphur `= 32 g mol^(-1)`).

Text Solution

Verified by Experts

The given values are:
`W_(B) = 3.24 g`, `W_(A) = 40 g`
`DeltaT_(b) = 0.81 K`, `K_(b) = 2.53 Kg mol^(-1)`
Using formula, `Mw_(B) = (K_(b) xx 1000 xx W_(B)) /(DeltaT_(b) xx W_(A))`
On substituting all the values, we get
`:. Mw_(B) = (2.53 xx 1000xx 3.24) / (0.81 xx 40) = 253`
Let the molecular formula of sulphur = `S_(x)`
Atomic mass of sulphur = `32`
Molecular mass = `32 xx x`
`:. 32x = 253`
`x= (253)/(32) = 7.91 ~~ 8`
`:.` Molecular formula of sulphur `= S_(8)`
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On dissolving 3.24 g of sulphur in 40 g of benzene, boiling point of solution 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 g mol^(-1) )

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 ).

Knowledge Check

  • On dissolving 3.24 g of sulphur in 40 g of benezene by 0.81 K. K_(b) value of benene is 2.53 k kg mol ^(-1) . Atomic mass of sulphur is 32 g mol ^(-1) . The molecular formula of sulphur is ______.

    A
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    B
    `S_(7)`
    C
    `S_(8)`
    D
    `S_(9)`

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