Chapter 14.2, Problem 20PP

Interpretation Introduction

Interpretation:

Amount of CaCl₂ is to be calculated to prepare 1.0 L of 0.10 M CaCl₂ solution.

Concept Introduction:

The concentrations of solutions are expressed in different ways. All of them are involved in the quantity of solute and the quantity of solution or solvent. Among them, molarity is widely used to express the concentration of a solution. Molarity: Molarity is defined as the number of moles of solute present in one liter of the solution.

In expression, $\text{Molarity}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{\text{V}\text{o}\text{l}\text{u}\text{m}\text{e}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Again $\text{Number}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{\text{m}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}$

Answer to Problem 20PP

11.1gm of CaCl₂ is needed to make 1.0L of 0.10M CaCl₂ solution.

Explanation of Solution

Given information:

The molarity of CaCl₂ solution = 0.10

Volume of CaCl₂ solution = 1.0 L

In expression, $\text{Molarity}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{\text{V}\text{o}\text{l}\text{u}\text{m}\text{e}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

  $\text{Molarity}\text{o}\text{f}\text{C}\text{a}\text{C}{\text{l}}_2\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\frac{\text{N}\text{u}\text{m}\text{b}\text{e}\text{r}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}{\text{CaCl}}_2}{\text{V}\text{o}\text{l}\text{u}\text{m}\text{e}\text{o}\text{f}{\text{CaCl}}_2\text{solution}(\text{i}\text{n}\text{l}\text{i}\text{t}\text{r}\text{e})}$

Number of moles of CaCl₂ = 0.10 M × 1.0 L

           = 0.10 (mol/litre) × 1.0 litre = 0.10

Again $\text{Number}\text{o}\text{f}\text{m}\text{o}\text{l}\text{e}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}{\text{m}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}$

  $=\frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{C}\text{a}\text{C}{\text{l}}_2}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}\text{o}\text{f}\text{C}\text{a}\text{C}{\text{l}}_2}$

  = 0.10

i.e, Mass of CaCl₂ = 110.98 (gm/mol) × 0.10 mol (Molecular weight of CaCl₂ = 110.98 gm/mol)

        = 11.1 gm