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

Sodium sulfate, #"Na"_2"SO"_4#, will dissociate in aqueous solution to form sodium cations, #"Na"^(+)#, and sulfate anions, #"SO"_4^(2-)#

#"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 #3#.

The equation for boiling-point elevation looks like this

#DeltaT_b = i * K_b * b#, where

#i# - the van't Hoff factor;
#K_b# - the ebullioscopic constant;
#b# - the molality of the solution.
#DeltaT_b# - the poiling point elevation - defined as #T_"b" - T_"b"^@#

The ebullioscopic constant of water is equal to #0.512^@"C kg mol"^(-1)#

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 #"29.7-g"# sample

#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#, the boiling-point elevation of the solution

#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

#T_"b"# - the boiling point of the pure solvent
#T_"b"^@# - the boiling point of the solution

#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")#