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

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

Step-by-step 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 = (mole \ of \ solute )/(kg \ of \ solvent )`

which can now be re-written as :

`molality = (mole \ of \ Ca(NO_3)_2)/(kg \ of \ water)`

`molality = ((mass \ of \ \ Ca(NO_3)_2)/(molar \ mass of \ Ca(NO_3)_2) )/(kg \ of \ water)`

`molality = ((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 * (1.86 \ ^0C/m) * (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