- #1
amcavoy
- 665
- 0
At 25oC the vapor pressure of pure water is 23.76 mmHg and that of seawater is 22.98 mmHg. Assuming that seawater contains only NaCl, estimate its concentration in molality units.
[tex]X_1=\frac{n_1}{n_1+n_2}\implies n_2=\frac{n_1-X_1n_1}{X_1}[/tex]
where n1 is the moles of solvent and n2 is the moles of solute.
[tex]22.98=X_1\left(23.76\right)\implies X_1=.9672[/tex]
and 1000 g of water is equal to 55.49 mol (n1), so plugging this all in gives:
[tex]n_2=1.88\text{mol}[/tex]
which would be the same as the molarity.
However, my textbook says that the molarity is .920 m. Where did I go wrong? Thanks a lot.