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 = 32 g mol^{−1}]

Advertisement Remove all ads

#### 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

âˆ†Tf = iKfm

0.383= i×3.83×0.8

i = 1/8

Hence, 8 sulphur atoms are undergoing association, as shown below:

8 S â‡Œ S8

Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads