# What is the freezing point of a solution prepared by dissolving 11.3 g of #Ca(NO_3)_2# (formula weight = 164 g/mol) in 115 g of water?

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The molal freezing point depression constant for water is 1.86° C/m.

The molal freezing point depression constant for water is 1.86° C/m.

##### 1 Answer

#### Explanation:

In order to find the freezing point of this solution, you must first find the **freezing-point depression** associated with dissolving that much calcium nitrate,

*Freezing-point depression* is a colligative property, which means that it depends on the **number of particles** of solute present in solution, and not on the nature of those particles.

Calcium nitrate is a **soluble** ionic compound that dissociates completely in aqueous solution to form

#"Ca"("NO"_3)_text(2(s]) -> "Ca"_text((aq])^(2+) + 2"NO"_text(3(aq])^(-)#

As you an see, *every mole* of calcium nitrate will produce **three moles** of ions in solution, **one mole** of calcium cations and **two moles** of nitrate anions.

Keep this in mind.

Now, the equation that describes *freezing-point depression* looks like this

#color(blue)(DeltaT_f = i * K_f * b)" "# , where

*van't Hoff factor*

*cryoscopic constant* of the solvent;

In your case, the cryoscopic constant of water is said to be equal to

#K_f = 1.86^@"C"/"m" = 1.86^@"C kg mol"^(-1)#

Now, the key here is the value of the *van't Hoff factor*, which tells you how many moles of particles are produced in solution *per mole* of dissolved solute.

Since we've established that one mole of calcium nitrate produces three moles of ions in aqueous solution, the van't Hoff factor will be

#i = 3#

To find the molality of the solution, you need to know two things

thenumber of molesof solutethe mass of the solvent expressed inkilograms

The number of moles of solute can be determined by using the given **formula weight**

#11.3 color(red)(cancel(color(black)("g"))) * ("1 mole Ca"("NO"_3)_2)/(164color(red)(cancel(color(black)("g")))) = "0.0689 moles Ca"("NO"_3)_2#

The molality of the solution will thus be

#b = "0.0689 moles"/(115 * 10^(-3)"kg") = "0.599 mol kg"^(-1)#

Now plug all this into the equation for freezing-point depression and get the value of

#DeltaT_f = 3 * 1.86^@"C" color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 0.599color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1)))) = 3.34^@"C"#

The freezing point of the solution will thus be

#color(blue)(DeltaT_f = T_f^0 - T_f)" " implies" "T_f = T_f^@ - DeltaT_f#

Here *pure solvent**.

In your case, you will have

#T_f = 0^@"C" - 3.34^@"C" = color(green)(-3.34^@"C")#

The answer is rounded to three sig figs.