Chapter 16, Problem 85A

(a)

Interpretation Introduction

Interpretation: The amount of $\text{KCl}$ (in grams) dissolved is to be calculated when $50.0\text{g}$ of $\text{KCl}$ is dissolved in $130\text{g}$ of water.

Concept Introduction: Solubility is a term defined as the maximum amount of solute particles that can be dissolved by a particular solvent for making a solution.

Answer to Problem 85A

The amount of $\text{KCl}$ , potassium chloride dissolved is $44.2\text{g}$ .

Explanation of Solution

The solubility of $\text{KCl}$ in $100\text{g}$ of water is $34.0\text{g}$ . This means that $100\text{g}$ of the water contains $34.0\text{g}$ $\text{KCl}$ .

Therefore, $1.0\text{g}$ of water contains $\frac{34}{100}\text{g}$ of $\text{KCl}$ .

For $130\text{g}$ of water, the amount of $\text{KCl}$ , potassium chloride is,

  $\frac{34}{100}\text{g}\times 130\text{g}=44.2\text{g}$ =34/100×130g

So, the dissolved amount of $\text{KCl}$ is $44.2\text{g}$ .

(b)

Interpretation Introduction

Interpretation: The amount of solute that came out of the solution when $50.0\text{g}$ of $\text{KCl}$ is dissolved in $130\text{g}$ of water is to be calculated.

Concept Introduction: The maximum amount of solute that can be dissolved by a solvent is solubility and the amount that cannot be dissolved by the solvent will come out as a precipitate in the solution.

Answer to Problem 85A

The amount of solute that came out or precipitate out of the solution is $5.8\text{g}$ .

Explanation of Solution

The amount of solute that can be dissolved in $130\text{g}$ of water is $44.2\text{g}$ .

So, the amount of $\text{KCl}$ precipitated can be calculated by subtracting the dissolved amount of solute from the total amount of solute particles.

That is,

The amount of $\text{KCl}$ came out is,

  $50-44.2=5.8\text{g}$ =50-44.2=5.8g

So, $5.8\text{g}$ of solute or $\text{KCl}$ came out of the solution.