What is the boiling point of a solution made by dissolving 31 g of NaCl in 559 g of water?

1 Answer
Nov 28, 2015

Answer:

#101^@"C"#

Explanation:

The important thing to recognize here is that sodium chloride is an electrolyte, which means that it will dissociate in aqueous solution to give sodium cations, #"Na"^(+)#, and chloride anions, #"Cl"^(-)#

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

This means that one mole of sodium chloride will produce two moles of ions in solution, one mole of sodium cations and one mole of chloride anions. Keep this in mind.

The equation for boiling-point elevation looks like this

#color(blue)(DeltaT_b = i * K_b * b)" "#, where

#i# - the van't Hoff factor;
#K_b# - the ebullioscopic constant of the solvent;
#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

To determine the molality of the solution, use sodium chloride's molar mass to help you find the number of moles you have in that #"31-g"# sample.

#31color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "0.5305 moles NaCl"#

This means that the molality of the solution will be - do not forget to convert the mass of the solvent from grams to kilograms

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

#b = "0.5305 moles"/(559 * 10^(-3)"kg") = "0.949 molal"#

Now, the van't Hoff factor, #i#, will be equal to #2# because you get two moles of particles per mole of sodium chloride added to the solution.

The boiling-point elevation will thus be

#DeltaT_"b" = 2 * 0.512^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.949 color(red)(cancel(color(black)("moles"))) color(red)(cancel(color(black)("kg"^(-1)))) = 0.97^@"C"#

This means that the boiling point of the solution will be

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

#T_"b" = 100^@"C" + 0.97^@"C" = color(green)(101^@"C")#