Question

What type of solution will be created when $35g$ of $N{H_4}Cl$ is added to $25ml$ of water at $80^\circ C$?

Answer 1

A saturated solution is a solution that contains the most extreme measure of solute that is equipped for being dissolved. At $20^\circ C$, the greatest measure of $NaCl$ that will dissolve in $100g$ of water is 36.0 g. If any more $NaCl$ is added past that point, it won't dissolve on the grounds that the solution is immersed. Imagine a scenario in which more water is added to the solution. Presently more $NaCl$ would be equipped for dissolving in the extra dissolvable.

Complete step-by-step answer: When we add more solute than a solvent can dissolve at a specific temperature, we'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 $65.6g$ per $100g$ of water at $80^\circ C$. This means that, if we add more than $65.6g$ of ammonium chloride to approximately $100ml$ of water, we'll get a saturated solution.
At that temperature, the density of water is $0.9718g/ml$(we know this)
$\rho = \dfrac{m}{v} \Rightarrow m = \rho .v = 0.9718\dfrac{g}{{ml}}.25ml = 24.3g$
This much water can dissolve
$24.3g.\dfrac{{65.5gN{H_4}Cl}}{{100.0g}} = 15.94gN{H_4}Cl$
Since we've added $35g$ of ammonium chloride to the solution, we've created a saturated solution. Our solution will contain an undissolved mass of $N{H_4}Cl$ equal to
${m_{undissolved}} = 35 - 15.94 = 19.1g$

Note: An unsaturated solution is a chemical solution wherein the solute concentration is lower than its equilibrium solvency. The entirety of the solutes dissolved in the dissolvable. At the point when a solute (frequently a solid) is added to a dissolvable (regularly a fluid), two cycles happen at the same time. One is dissolution, another one is crystallization.