# What is the maximum number of grams of "NH"_4"Cl" that will dissolve in 200 grams of water at 70°C?

Jul 5, 2016

$\text{120 g NH"_4"Cl}$

#### Explanation:

Your tool of choice here will be the solubility graph for ammonium chloride, $\text{NH"_4"Cl}$, which looks like this

The solubility graph provides you with information on the solubility of a given salt at various temperatures. In this case, the graph shows you the solubility of ammonium chloride per $\text{100 g}$ of water at temperatures ranging from ${0}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$.

The actual curve represents the solubility of the salt, i.e. the maximum amount that can be dissolved in $\text{100 g}$ of water to form a saturated solution.

At ${70}^{\circ} \text{C}$, ammonium chloride has a solubility of about $\text{62 g / 100 g H"_2"O}$. This tells you that can only hope to dissolve $\text{62 g}$ of ammonium chloride for every $\text{100 g}$ of water before the solution becomes saturated.

After this point, any extra salt added would remain undissolved.

So, you know how much salt can be dissolved in $\text{100 g}$ of water, so use this as a conversion factor to find the solubility of the salt in $\text{200 g}$ of water

200 color(red)(cancel(color(black)("g H"_2"O"))) * ("62 g NH"_4"Cl")/(100color(red)(cancel(color(black)("g H"_2"O")))) = "124 g NH"_4"Cl"

I will leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the mass of water

"solubility at 70"""^@"C in 200 g H"_2"O" = color(green)(|bar(ul(color(white)(a/a)color(black)("120 g")color(white)(a/a)|)))