Question

400 g of MgCl2 is dissolved in 1000 g of water. What is the freezing point of this solution? Kf.p.= 1.86 degrees C/m for water

Answer 1

As solvet wt is 1000g of water, we will use molal depression constant equation as:

Kf = Depression in freezing point x Wt of solvent (W) x molecular wt of solute (m) / 1000 x wt of solute (w)

Where as Kf = molal depression constant, molecular wt of MgCl2 = 95.21

and Depression in freezing point = T_solvent - T_solution

So from question Depression in freezing point = Kf x 1000 x w / W x m

                                                                      = 1.86 x 1000 x 400 / 1000 x 95.21

                          Depression in freezing point    = 7.81 ^oC

So freezing point of solution T_solution = 0 - 7.81

ie.    T_solution = -7.81 ^oC