0
$\begingroup$

Here's a challenge problem I created! None of my students could get it; can you?

$25$ grams of table salt (NaCl) is dissolved in $150$ grams of water. The temperature of the solution is raised to its boiling point until sodium chloride begins to precipitate.

How much water has boiled off when precipitation begins? Also, what would be the temperature of the boiling solution when precipitation begins?

Solubility Table

$\endgroup$

closed as off-topic by Loong, Klaus-Dieter Warzecha, jerepierre, ron, Philipp Feb 24 '15 at 0:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

0
$\begingroup$

One expects 25 g of $\ce{NaCl}$ to saturate 62.5 g of water at 100 °C. However, 25 g of $\ce{NaCl}$, 58.443 g/mol, is 0.4278 mol in 62.5 ml of water at 100 °C. Colligative property boiling point elevation is +0.52 (°C kg)/mol. 0.8556 mol of particles in 62.5 ml of water is 13.69 mol/kg or 7.1 °C elevated.

$\ce{NaCl}$ solubility/temp increases from about 36 g/100 g at 0 °C to 40 g/100 g at 100 °C or 0.04 g/°C. This gives first approximation 0.2847 g more solubility. "25.28" g $\ce{NaCl}$ in 62.5 g water at 107 °C is then 25 g $\ce{NaCl}$ in 61.8 g water. 88.2 g water boiled off (with some decimal trim for Zeno’s paradox).

Repeat the problem with ammonium cyanate.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.