(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?

Explanation

(a) `W_(B) = 2.56g`
`W_(A) = 100g`,
`DeltaT_(f) = 0.383 K`
`K_(f)=3.83 K kg//mol`
`DeltaT_(f) = Kf m`
`DeltaT_(f) = (Kf xx W_(B) xx 1000)/(M_(B) xx W_(A))`
`0.383 = (3.83 xx 2.56 xx 1000)/(M_(B) xx 100)`

`M_(B) = 256`
Sx = 256
`x xx 32 = 256`
`x = (256)/(32) = 8`
Hence, sulphur has `S_(8)` form.
(b) Blood cells are isotonic with 0.9% NaCl solution :
(i) Blood cell placed in 1.2% NaCl solution (Hypertonic Solution) : Blood cell will shrink (high to low concentration)
(ii) Blood cell places in 0.4% NaCl solution (Hypotonic Solution) - Blood cell will swell and even burst (low to high concentration).