What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of water? The molal freezing point depression constant for water is 1.86 °C/m.

freezing point   (°C) of the solution =  - 3.34° C

Explanation:

From the given information:

The freezing point (°C) of a solution can be prepared by using the formula:

where;

i = vant Hoff factor

the vant Hoff factor is the totality of the number of ions in the solution

Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3  

= 1.86 °C/m

m = molality of the solution and it can be determined by using the formula

which can now be re-written as :

molality = 0.599 m

∴

The freezing point (°C) of a solution can be prepared by using the formula:

the freezing point of water - freezing point of the solution

3.34° C = 0° C -  freezing point of the solution

freezing point  (°C) of the solution =  0° C - 3.34° C

freezing point   (°C) of the solution =  - 3.34° C

i have no clue hope you find the answer