An aqueous solution containing `NH_(4)Cl` completely ionised in water freezes at `-0.3272^(@)C. K_(f)` of water is `1.86 K "molality"^(-1)`. The solution on heating to `50^(@)C` gives a collection to dry gases which register `2.463 atm` pressreu in `1` litre container at `27^(@)C`. Assuming no water vaporises at `50^(@)C`, calculate the temperature at which the solution on cooling back freezes out.

Explanation

`Delta T_(f) = K_(f) xx "Molality xx (1+alpha)`
`0.372 = 1.86 xx "Molality xx 2`
`:."Molality" = 0.1`
Thus `0.1` mole of `NH_(4)Cl` is present in `1 kg` water. On heating `NH_(4)Cl` decomposes to give `NH_(3)` and `HCl`.
Thus, `P_(T) = P_(NH_(3)) + P_(HCl) = 2.463`
`:. PV = nRT`
`2.463 xx 1 = n xx 0.0821 xx 300`
`:. n = 0.1`
Thus `{:(NH_(4)Cl(s),hArr,NH_(3)(g)+,HCl_(g),),(0.1,,0,0,),((0.1-x),,x,x,):}`
`because 2x = 0.1`
`:. x = 0.05`
Moles of `NH_(4)Cl` left in sol. `= 0.1 - 0.05 = 0.05`
`:. Delta T_(f) = 1.86 xx 0.05 xx 2`
`= 0.186 K`