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Question
When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx).
(Kf for CS2 = 3.83 K kg mol−1, Atomic mass of sulphur = 32 g mol−1]
Solution
Given:
Kf = 3.83 K kg mol−1
Mass of solute = 2.56 g
Mass of solvent = 100 g
Therefore,
`"Molality of the solution, "m =2.56/32xx1000/100=0.8m`
The depression in freezing point of a solution is given as
∆Tf = iKfm
0.383= i×3.83×0.8
i = 1/8
Hence, 8 sulphur atoms are undergoing association, as shown below:
8 S ⇌ S8
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