On dissolving 3.24 g of sulphur in ...

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 ______.

`s_(6)`

`S_(7)`

`S_(8)`

`S_(9)`

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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 ).

5g sulphur is present in 100 g of CS_(2). DeltaT_(b) of solution 0.954 and K_(b) is 4.88. The molecular formula of sulphur is

(a) When 2.56 g of sulphur was dissolved in 100 g of CS_(2) , the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S_(x)) . ( K_(f) for CS_(2) = 3.83 K kg mol^(-1) , Atomic mass of sulphur = 32g mol^(-1) ] (b) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing. (i) 1.2% sodium chloride solution? (ii) 0.4% sodium chloride solution?

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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