Question

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.

Answer 1

Answer:

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:

$\Delta T = iK_fm$

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  

$K_f$ = 1.86 °C/m

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

$molality = \dfrac{mole \ of \ solute }{kg \ of \ solvent }$

which can now be re-written as :

$molality = \dfrac{mole \ of \ Ca(NO_3)_2}{kg \ of \ water}$

$molality = \dfrac{\dfrac{mass \ of \ \ Ca(NO_3)_2}{molar \ mass of \ Ca(NO_3)_2} }{kg \ of \ water}$

$molality = \dfrac{\dfrac{11.3 \ g }{164 \ g/mol} }{0.115 \ kg }$

molality = 0.599 m

∴

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

$\Delta T = iK_fm$

$\Delta T =3 \times (1.86 \ ^0C/m) \times (0.599 \ m)$

$\Delta T =3.34^0 \ C$

$\Delta T =$ 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