Question

What is the freezing point of a solution of 155 g LiCl in 575 g H20?

Answer 1

`-23.6°C`

Explanation:

Recall,

`DeltaT_f = iK_fm`

Where `i` is the number of species in solution, `m` is the molality of the species, and `K_f` is the freezing point constant for that particular solvent (water, in this case).

Because `"LiCl"` dissociates in water solution into `"Li"^+` and `"Cl"^(-)` ions, `i = 2`.

Hence,

`DeltaT_F = 2 * (1.86°C)/m * (155g * ("mol")/(42.44g))/(575g * (kg)/(10^3g))`

`therefore T_f approx 23.6°C`

and by extension `-23.6°C` is the freezing point of that solution given your data.