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

Jul 31, 2017

We need the $\text{molal boiling point elevation constant}$ for water.........and thus the boiling point of the solution is $100.33$ ""^@C.

#### Explanation:

This site gives ${K}_{B} = 0.512 \cdot K \cdot k g \cdot m o {l}^{-} 1$ as the so-called $\text{ebullioscopic constant}$ 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 $0.321 \cdot m o l a l$ with respect to BOTH $N {a}^{+}$ and $C {l}^{-}$ ions.......

Now $\Delta {T}_{b} = {K}_{B} \times C \times i$, where $i$ is the number of ions in solution......

$= 0.512 \cdot K \cdot k g \cdot m o {l}^{-} 1 \times 0.321 \cdot m o l \cdot k {g}^{-} 1 \times 2$

$= 0.33 \cdot K$; 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.