Question

The freezing point depression constant for water is 1.86 oC/mol of particles. If 345 g of calcium chloride (CaCl2) salt is dissolved in 1000 g of water, what will be the freezing pointof the solution?

Answer 1

$\Delta$t_f=ik_fm

K_f=1.86

i= 1+(3-1)$\alpha$ .......... {$\alpha$ =1}

$\implies$ i = 3

m = $n_{solute}\over {m_{H_2O(kg)}}$ $\over$

m=$3.1\over 1 (kg)$ ............{n_solute= $345\over 111$ =3.1}

$\implies$ m= 3.1

Δt_f=$3\times1.86\times3.1$

=17.298

Answer = 0-17.298

= -17.298