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

1 Answer
anor277
Jul 31, 2017

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.

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