Chapter 4, Problem 4.22P

(a)

Interpretation Introduction

Interpretation:

The mass $\left(\text{g}\right)$ of solute in $185.8\text{ mL}$ of $0.267M$ calcium acetate is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol/L}$.

The expression to calculate the moles of the solute when molarity of solution and volume of solution are given is as follows:

$\text{Moles of solute}\left(\text{mol}\right)=\left[\begin{array}{c}\text{volume of solution}\left(\text{L}\right)\\ \left(\frac{\text{molarity}\text{of}\text{solution}\left(\text{mol}\right)}{\text{1 L of solution}}\right)\end{array}\right]$

The expression to calculate the amount of solute when moles of solute and molecular mass are given is as follows:

$\text{amount of solute}\left(\text{g}\right)=\text{moles of solute}\left(\text{mol}\right)\left(\frac{\text{molecular mass of solute}\left(\text{g}\right)}{1\text{mole of solute}}\right)$

(b)

Interpretation Introduction

Interpretation:

The molarity of $500\text{ mL}$ of solution containing $21.1\text{ g}$ of potassium iodide is to be determined.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol/L}$.

The expression to calculate the molarity of a solution when moles of solute and volume of solution are given is as follows:

$\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

The expression to calculate the moles of solute when given mass and molecular mass of solute are given is as follows:

$\text{Moles of solute}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of solute}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of solute}\left(\text{mol}\right)}{\text{molecular mass of solute}\left(\text{g}\right)}\right)\end{array}\right]$

(c)

Interpretation Introduction

Interpretation:

The amount $\left(\text{mol}\right)$ of solute in $145.6\text{ L}$ of $0.850M$ sodium cyanide is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol/L}$.

The expression to calculate the amount of compound in moles when molarity of solution and volume of solution is given is as follows:

$\text{moles of solute}\left(\text{mol}\right)=\left(\text{volume of solution}\left(\text{L}\right)\right)\left(\frac{\text{molarity of solution}\left(\text{mol}\right)}{\text{1}\text{L of solution}}\right)$