Question

Instead of using NaCI to melt the ice on your sidewalk, you decide to use CaCl2. If you add 34.7 g of CaClz to 150. g of water, what is the freezing point of the solution? (Ksp for water is -1.858 C/m.)

Answer 1

$$n(CaCl_2)=\frac{m}{M}=\frac{34.7}{110.98}=0.313 mol$$

$$molarity=\frac{n}{m(solute)}=\frac{0.313 mol}{0.150 kg}=2.09\frac{mol}{kg}$$

$$\Delta T_{fp}=k_f\times m=(-1.858\frac{C}{m})\times 2.09=-3.88 C$$