what type of solution will be created when 35 grams of Nh4Cl is added to 25 ml of water at 80 Celsius?

Apr 18, 2015

You will create a saturated solution.

When you add more solute than a solvent can dissolve at a specific temperature, you'll create a saturated solution. The solute will dissolve in the limit of its solubility; anything that exceeds that amount will remain undissolved in solution.

The solubility of ammonium chloride is listed at 65.6 per 100.0 g of water at ${80}^{\circ} \text{C}$. This means that, if you add more than 65.6 g of ammonium chloride to approximately 100 mL of water, you'll get a saturated solution.

At that temperature, the density of water is 0.9718 g/mL (http://antoine.frostburg.edu/chem/senese/javascript/water-density.html), which means that you'll get

$\rho = \frac{m}{v} \implies m = \rho \cdot V = 0.9718 \text{g"/cancel("mL") * 25cancel("mL") = "24.3 g}$

This much water can dissolve

24.3cancel("g") * ("65.5 g "NH_4Cl)/(100.0cancel("g")) = "15.94 g " $N {H}_{4} C l$

Since you've added 35 g of ammonium chloride to the solution, you've created a saturated solution. Your solution will contain an undissolved mass of $N {H}_{4} C l$ equal to

${m}_{\text{undissolved" = 35 - 15.94 = "19.1 g}}$