The addition of `3g` of a substance to `100g C CI_(4)(M = 154 g mol^(-1))` raise the boiling point of `C CI_(4)` by `0.60^(@)C` of `K_(b)(C CI_(4))` is `5.03 kg mol^(-1) K`. Calculate:
(a) the freezing point depression
(b) the relative lowering of vapour pressure
(c) the osmotic pressure at `298K`
the molar mass of the substance
Given `K_(f)(C CI_(4)) = 31.8 kg mol^(-1) K` and `rho` (density) of solution `= 1.64g//cm^(3)`

Answer

(a) `3.793^(@)C` (b) `0.018`, (c) `4.65 atm`, (d) `251.5`

Explanation

(a) `(Delta T_(t))/(Delta T_(b)) = (k_(f))/(k_(b))=Delta T_(f)=(0.6xx31.8)/(5.03)=3.793^(@)C`
(b) Relative lowering of vapour pressure =
`(n)/(n+N)=((3)/(251.5))/((3)/(251.5)+(100)/(154))=0.018`
(c ) `pi = CRT`
`n = (3)/(251.5)=0.012`
`v = (103)/(1.64) = 62.8 mL`
`pi = (0.012)/(0.0628)xx0.0821xx298=4.65` atm
(d) `0.6 = (5.03xx3xx1000)/(M_(W)xx100) = gt M_(W) = 251.5`