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

Hint: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.