# What is the boiling point of an ionic solution containing 29.7 g #Na_2SO_4# and 84.4 g water, assuming 100% ionization?

##### 1 Answer

#### Answer:

#### Explanation:

*Sodium sulfate*,

#"Na"_2"SO"_text(4(aq]) -> 2"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#

Notice that **every formula unit** of sodium sulfate produces **three ions** in solution, tow sodium cations and one sulfate anion. This means that the solution's **van't Hoff factor** will be equal to

The equation for *boiling-point elevation* looks like this

**ebullioscopic constant**;

The *ebullioscopic constant* of water is equal to

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

Now, the molality of the solution is defined as the number of moles of solute, which in your case is sodium sulfate, divided by the mass of the solvent, which in your case is water, expressed in **kilograms**.

Use sodium sulfate's molar mass to determine how many moles you have in that

#29.7color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"SO"_4)/(142.04color(red)(cancel(color(black)("g")))) = "0.2091 moles Na"_2"SO"_4#

The molality of the solution will thus be

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

#b = "0.2091 moles"/(84.4 * 10^(-3)"kg") = "2.477 moles kg"^(-1) = "2.477 molal"#

Now plug in your values and solve for

#DeltaT_b = 3 * 0.512^@"C"color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 2.477color(red)(cancel(color(black)("moles")))color(red)(cancel(color(black)("kg"^(-1))))#

#DeltaT_b = 3.805^@"C"#

The boiling point of the solution will thus be

#DeltaT_"b" = T_"b" - T_"b"^@" "# , where

**pure solvent**

#T_"b" = DeltaT_"b" + T_"b"^@#

#T_"b" = 3.805^@"C" + 100^@"C" = 103.805^@"C"#

Rounded to three sig figs, the answer will be

#T_"b" = color(green)(104^@"C")#