# Question #9d5e0

##### 1 Answer

#### Explanation:

The first thing to do here is calculate the **freezing-point depression**,

Once you know that, you can use the freezing point of the **pure solvent** to find the freezing point of the solution.

So, the freezing-point depression is calculated using the equation

#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;

The cryoscopic constant of water is equal to

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

Now, sodium chloride, **soluble** ionic compound, which means that is dissociates **completely** in aqueous solution to form sodium cations and chloride anions

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

Notice that **every mole** of sodium chloride added to the solution produces **mole** of sodium cations and **mole** of chloride anions.

This means that the *van't Hoff factor* for sodium chloride, which essentially tells you how many moles of particles of solute are produced in solution **per mole** of solute dissolved, will be equal to

Plug in your values to find the freezing-point depression of the solution

#DeltaT_f = 2 * 1.86""^@"C" color(red)(cancel(color(black)("kg"))) * color(red)(cancel(color(black)("mol"^(-1)))) * 0.50color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

#DeltaT_f = 1.86^@"C"#

The freezing-point depression is defined as

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

Here

**pure solvent**

Pure water freezes at

#DeltaT_"f sol" = 0^@"C" - 1.86^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(-1.9^@"C")color(white)(a/a)|)))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the molality of the solution.