# What is the freezing point of a solution that contains 0.5 moles of Nal in 500g of water? (Kf = 1.86C/m; molar mass of water = 18g)

##### 1 Answer

#### Explanation:

Your strategy here will be to

determine the,van't Hoff factorfor sodium iodide#"NaI"# calculate themolalityof the solutioncalculate thefreezing-point depressionof the solution

The idea is that the freezing point of a *solution* is **lower** than the freezing point of the *pure solvent*, which for water is

Now, the difference between the freezing point of the pure solvent and the freezing point of the solution is given by the **freezing-point depression**, which can be calculated using the equation

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

Here

*van't Hoff factor*

*cryoscopic constant* of the solvent;

The problem provides you with the cryoscopic constant for water

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

Now, sodium iodide is **soluble** in aqueous solution, which means that it dissociates completely to form sodium cations,

#"NaI"_ ((aq)) -> "Na"_ ((aq))^(+) + "I"_ ((aq))^(-)#

Notice that **every mole** of sodium iodide that is dissolved in solution produces **two moles** of particles of solute, i.e. ions.

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

#i = 2 -># one moleof solute dissolved,two molesof ions produced

The **molality** of the solution is defined as the number of moles of solute present in **one kilogram of solvent**. Your solution contains **moles** of sodium iodide and

#500 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.5 kg"#

of water, your **solvent**. This means that the molaity of the solution will be

#b = "0.5 moles"/"0.5 kg" = "1 mol kg"^(-1)#

You now have what you need to calculate the freezing-point depression

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

#DeltaT_f = 3.72^@"C"#

The freezing point of the *solution* will thus be

#DeltaT_"f sol" = 0^@"C" - 3.72^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(-3.7^@"C")color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that your values only justify one sig fig for freezing point of the solution.