What is the boiling point of a 0.321m aqueous solution of #NaCl#?

1 Answer
Jul 31, 2017

Answer:

We need the #"molal boiling point elevation constant"# for water.........and thus the boiling point of the solution is #100.33# #""^@C#.

Explanation:

This site gives #K_B=0.512*K*kg*mol^-1# as the so-called #"ebullioscopic constant"# for water. Of course there is a catch, and this requires us to consider ALL the species in solution, i.e. in the given solution we have concentrations of #0.321*molal# with respect to BOTH #Na^+# and #Cl^-# ions.......

Now #DeltaT_b=K_BxxCxxi#, where #i# is the number of ions in solution......

#=0.512*K*kg*mol^-1xx0.321*mol*kg^-1xx2#

#=0.33*K#; and this is a boiling point elevation with respect to the PURE solvent......And thus the boiling point of the solution is #100.33# #""^@C#.