# A solution is made by dissolving 25.0 g of magnesium chloride crystals in 1000 g of water. What will be the freezing point of the new solution assuming complete dissociation of the #MgCl_2# salt?

##
What will the boiling point of the new solution assuming complete dissociation of the #MgCl_2# salt?

What will the boiling point of the new solution assuming complete dissociation of the

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

**!! Long ANSWER !!**

*Magnesium chloride*, **soluble** ionic compound that dissociates in aqueous solution to form magnesium cations,

#"MgCl"_text(2(aq]) -> "Mg"_text((aq])^(2+) + 2"Cl"_text((aq])^(-)#

Now, if you assume that the salt *dissociates completely*, you can say that **every mole** of magnesium chloride will produce **three moles** of ions in solution

one moleof magnesium cationstwo molesof chloride anions

As you know, the **freezing point** of a solution depends on *how many* particles of solute you have present, not on the nature of the solute

Mathematically, you can express the *freezing-point depression* of a solution by using the equation

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

*van't Hoff factor*

*cryoscopic constant* of the solvent;

The cryoscopic constant of water is equal to

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

The **van't Hoff factor** tells you the ratio between the concentration of particles produces in solution when a substance is dissolved, and the concentration of said substance.

Since you know that every mole of magnesium chloride produces **three moles** of ions in solution, you can say that the van't Hoff factor will be equal to

In order to find the molality of the solution, you need to know how many **moles** of solute you have in that

To do that, use the compound's **molar mass**

#25.0 color(red)(cancel(color(black)("g"))) * "1 mole MgCl"_2/(95.21color(red)(cancel(color(black)("g")))) = "0.2626 moles MgCl"_2#

Now, molality is defined as moles of solute per **kilograms** of solvent.

#color(blue)(b = n_"solute"/m_"solvent")#

In your case, you will have

#b = "0.2626 moles"/(1000 * 10^(-3)"kg") = "0.2626 molal"#

This means that the freezing-point depression will be

#DeltaT_f = 3 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2626color(red)(cancel(color(black)("mol")))color(red)(cancel(color(black)("kg"^(-1)))) = 1.465^@"C"#

The freezing-point depression is defined as

#color(blue)(DeltaT_f = T_f^@ - T_"f sol")" "# , where

**pure solvent**

This means that you have

#T_"f sol" = T_f^@ - DeltaT_f#

#T_"f sol" = 0^@"C" - 1.465^@"C" = -1.465^@"C"#

You *should* round this off to one sig fig, since that is how many sig figs you have for the mass of water, but I'll leave it rounded to two sig figs

#T_"f sol" = color(green)(-1.5^@"C")#

Now for the boiling point of this solution. The equation for *boiling-point elevation* looks like this

#color(blue)(DeltaT_b = i * K_b * b)" "# , where

*van't Hoff factor*

The *ebullioscopic constant* for water is equal to

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

Plug in your values to get

#DeltaT_b = 3 * 0.512^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2626 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1)))) = 0.403^@"C"#

The boiling-point elevation is defined as

#color(blue)(DeltaT_b = T_"b sol" - T_b^@)" "# , where

**pure solvent**

In your case, you have

#T_"b sol" = DeltaT_b + T_b^@#

#T_"b sol" = 0.403^@"C" + 100^@"C" = 100.403^@"C"#

I'll leave this answer as

#T_"b 'sol" = color(green)(100.4^@"C")#