# Salt, NaCl, has a solubility of 35.7g per 100g of water at 0 degrees C. What is the lowest possible melting point for ice that could be obtained considering the solubility of NaCl in water? The Kf for H2O is 1.86 C/m

##### 1 Answer

#### Answer:

#### Explanation:

Your tool of choice here will be the equation that expresses the **freezing-point depression** of a solution, which looks like this

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = i * K_f * bcolor(white)(a/a)|)))" "# , where

*van't Hoff factor*

*cryoscopic constant* of the solvent;

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

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

So, the idea here is that you can only dissolve

The thing to remember about sodium chloride is that the compound is a **strong electrolyte**, which means that it dissociates completely in aqueous solution to form sodium cations,

#"NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl"^(-)#

This means that the *van't Hoff factor*, which tells you the ratio that exists between the number of moles of solute that are *dissolved* and the number of moles of **solute particles** produced in solution, will be equal to

#i = 2#

That is the case because **one mole** of sodium chloride will dissociate completely to form **one mole** of sodium cations and **one mole** of chloride anions, hence **two moles** of particles.

Now, let's assume that you're working with **molar mass** to determine how many *moles* of sodium chloride you'd get in that

#35.7 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.443 color(red)(cancel(color(black)("g")))) = "0.61085 moles NaCl"#

The **molality** of a solution tells you how many moles of solute you get **per kilogram of solvent**

#color(blue)(|bar(ul(color(white)(a/a)"molality" = "moles of solute"/"kilogram of solvent"color(white)(a/a)|)))#

In your case, you will have

#b = "0.61085 moles"/(100 * 10^(-3)"kg") = "6.1085 mol kg"^(-1)#

The freezing-point depression of this solution would be

#DeltaT_f = 2 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 6.1085 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

#DeltaT_f = 22.72^@"C"#

The freezing-point depression is defined as

#color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" "# , where

**pure solvent**

The freezing point of the solution will thus be

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

#T_"f sol" = 0^@"C" - 22.72^@"C" = color(green)(|bar(ul(color(white)(a/a)-22.7^@"C"color(white)(a/a)|)))#

I'll leave the answer rounded to three **sig figs**.