Question

Potassium chloride has a solubility of $43gofKClin100.g$of $H_{2}Oat50°C.$Determine if each of the following forms an unsaturated or saturated solution at $20°C:(9.3)$

a) adding localid="1653998984557" $25gofKClto100.gof{H}_{2}O$

b) adding localid="1653998991946" $15gofKClto25gof{H}_{2}O$

c) addinglocalid="1653998999630" $86gofKClto150.gof{H}_{2}O$

Short answer

Part a) As a corollary, this solution is unsaturated.

Part b) As a conclusion, this is a Super Saturate solution.

Part b) As a function, this is a Super Saturate solution.

Step-by-step solution

Introduction (Part a)

Saturate solution: A saturated solution is one that has the greatest levels of dissolved solute at equilibrium.

A solution with a lower quantity than a saturated solution. It will be capable of dissolving a higher amount of solute.

Solution for super-saturation: A solution with a stronger presence of solute solute than a saturated solution.

Given data (Part a).

To $50°C$, $KCl$has a solubility of roughly localid="1653999113452" $\frac{43g}{100ml}$, or localid="1653999122657" $0.43gm{l}^{-1}$. The mixture is saturated if there is localid="1653999129676" $43gKCLin100ml$of water at $50°C$.

Explanation (Part a).

a. If $25gKCl$is added to $100g$of water, its solubility is $0.25gm{l}^{-1}$, which is below the soluble of $KCl$at localid="1653999155241" $50°C(0.43gm{l}^{-1})$, showing that the solution is unsaturated.

Given data (Part b).

Evaluate the number of $KCl$per ml in the following solution:

localid="1653999171490" $\frac{15g}{25ml}=0.6gm{l}^{-1}$

Explanation (Part b).

Because this value exceeds the solubility of $KCl;0.43gm{l}^{-1}$, this is a Super Saturate solution.

Given data (Part c). 

Evaluate the number of $KCl$per ml in the following solution:

localid="1653999194073" $\frac{86g}{150ml}=0.43gm{l}^{-1}$

Explanation (Part c).

So this value is higher the solubility of $KCl(0.43gm{l}^{-1})$, this is a Super Saturate solution.