Chapter 14, Problem 13STP

Interpretation Introduction

Interpretation:

Moles of KClO₃ is required to make 1 L of a saturated solution of KClO₃ at 75 °C has to be calculated from the graph.

Concept introduction:

Solubility: Solubility can be defined as maximum amount of solute that can dissolve in known amount of solvent at a given temperature and pressure. During solution formation, solute and solvent particles collide with each other, so that the solute gets dissolves in solvent. Solubility is affected by raising the temperature of the solvent because the kinetic energy of its particles is increased, resulting in more-frequent collisions and collisions with greater energy than those that occur at lower temperatures.

Answer to Problem 13STP

0.245 moles of KClO₃ is required to make 1 L of a saturated solution of KClO₃ at 75 °C.

In the graph, starts at 75 °C, go straight up until you hit the line of KClO₃. Then go across until hit the Y-axis. The mass it shows is 30 gm. So, at 75 °C the solubility of KClO₃ is 30 gm. Therefore 30 gm of KClO₃ can be dissolved in 100 gm water at 75 °C.

$\begin{array}{l}\mathrm{Moles}ofKCl{O}_3=\frac{MassofKCl{O}_3}{MolecularweightofKCl{O}_3}\\ \mathrm{Moles}ofKCl{O}_3=\frac{30gm}{122.55gm/mole}=0.245mole\end{array}$

0.245 moles of KClO₃ is required to make 1 L of a saturated solution of KClO₃ at 75 °C.

Explanation of Solution

In the graph, starts at 75 °C, go straight up until you hit the line of KClO₃. Then go across until hit the Y-axis. The mass it shows is 30 gm. So, at 75 °C the solubility of KClO₃ is 30 gm. Therefore 30 gm of KClO₃ can be dissolved in 100 gm water at 75 °C.

$\begin{array}{l}\mathrm{Moles}ofKCl{O}_3=\frac{MassofKCl{O}_3}{MolecularweightofKCl{O}_3}\\ \mathrm{Moles}ofKCl{O}_3=\frac{30gm}{122.55gm/mole}=0.245mole\end{array}$

0.245 moles of KClO₃ is required to make 1 L of a saturated solution of KClO₃ at 75 °C.