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)`3.79^(@)"C"`,(b)0.018, (c)4.65 atm, (d) 251.5

Explanation

(a)`0.60="K"_("b")xx"m"`
`Delta"T"_("f")="K"_("f")xx"m"`
`(Delta"T"_("f"))/0.60="K"_("f")/"K"_("b")implies31.8/5.03`
`Delta"T"_("f")=3.79^(@)"C"`
(b)`"m"=(Delta"T"_("b"))/"K"_("b")=0.60/5.03=0.12`
`"n"_("solute")=0.12," n"_("solvent")=1000/154=6.49`
Relative lowering of V.P. `"n"_("solute")/" n"_("solvent")=0.12/6.49=0.018`
(c) Amount of solvent in 100 gm of solvent =3gm
Amount of solute in 1000 gm of solvent=30gm
`"W"_("solution")=1030"gm," " V"_("solution")=1030/rho_("solution")=1030/1.64"cm"^(3)=1030/1640"litre"`
`pi=("n"_("solute")"RT")/"V"_("solution")`
`=(0.12xx0.082xx298)/(1030/1640)=4.65("atm")`
(d) `0.12=30/"m"_("solute")`
`"m"_("solute")=3000/0.12=250("app")`