Chapter 16, Problem 60A

(a)

Interpretation Introduction

Interpretation: The moles and grams of the solute $1.0\text{ L}$ of $0.50\text{ M}$

  $\text{NaCl}$ are to be calculated.

Concept Introduction: Mole concept is the method used to define the amount of a compound and the value of 1 mole is $6.022\times {10}^{23}$ Avogadro number.

The expression for moles of solute is as follows:

  $moles\text{ of solute}=Molarity\times Volume$

The expression for the mass of the compound is as follows:

  $Mass\text{of Compound}=Molar\text{ m}ass\times moles$

Answer to Problem 60A

The moles and mass of $\text{NaCl}$ are $0.5\text{ moles}$ and $29.25\text{ g}$ .

Explanation of Solution

Given,

Molar mass of $\text{NaCl}$ is $58.44\text{ g/mol}$ .

Volume of $\text{NaCl}$ is $1.0\text{ L}$ .

Molarity of $\text{NaCl}$ is $0.50\text{ M}$ .

Substitute, $0.50M$ for molarity and $1.0\text{ L}$ for the volume of $\text{ NaCl}$ in the expression for moles of solute.

  $\begin{array}{c}moles\text{of NaCl}=0.50\text{ M}\times 1.0\text{ L}\\ =0.5\text{ moles}\end{array}$

Substitute $58.44\text{ g/mol}$ for molar mass and $0.5\text{ moles}$ for the moles of $\text{ NaCl}$ in the expression for mass of the compound.

  $\begin{array}{c}Mass\text{of Compound}=58.5\text{ g/mol}\times 0.5moles\\ =29.25\text{ g}\end{array}$

The moles and mass of $\text{NaCl}$ are $0.5\text{ moles}$ and $29.25\text{ g}$ .

(b)

Interpretation Introduction

Interpretation: The moles and grams of the solute $5\times {10}^2\text{ mL}$ of $2.0\text{ M}$

  ${\text{KNO}}_3$ are to be calculated.

Concept Introduction: Mole concept is the method used to define the amount of a compound and the value of 1 mole is $6.022\times {10}^{23}$ Avogadro number.

Answer to Problem 60A

The moles and mass of ${\text{KNO}}_3$ are $1.0mole$ and $\text{1}\text{.0}\times {\text{10}}^2\text{ g}$ .

Explanation of Solution

Given, Molar mass of ${\text{KNO}}_3$ is $101.10\text{ g/mol}$ .

Volume of ${\text{KNO}}_3$ is $5\times {10}^2\text{ mL}$ .

Molarity of ${\text{KNO}}_3$ is $2.0\text{ M}$ .

Convert $\text{mL}$ to $\text{L}$ .

  $\begin{array}{c}5\times {10}^2\text{ mL}=5\times {10}^2\text{ mL}\times \frac{1\text{L}}{1000\text{mL}}\\ =0.5\text{L}\end{array}$

Substitute, $2.0\text{ M}$ for molarity and $0.5\text{ L}$ for the volume of ${\text{KNO}}_3$ in the expression for moles of solute.

  $\begin{array}{c}moles{\text{of KNO}}_3=0.5\text{ L}\times 2.0\text{ M }\\ =1.0\text{ moles}\end{array}$

Substitute, $101.10\text{ g/mol}$ for molar mass and $1.0\text{ moles}$ for the moles of ${\text{KNO}}_3$ in the expression for mass of the compound.

  $\begin{array}{c}Mass\text{of Compound}=101.10\text{ g/mol}\times 1.0moles\\ =101.1\text{ g}\\ \approx \text{1}\text{.0}\times {\text{10}}^2\text{ g}\end{array}$

The moles and mass of ${\text{KNO}}_3$ are $1.0mole$ and $\text{1}\text{.0}\times {\text{10}}^2\text{ g}$ .

(c)

Interpretation Introduction

Interpretation: The moles and grams of the solute $250\text{ mL}$ of $0.10\text{ M}$

  ${\text{CaCl}}_2$ are to be calculated.

Concept Introduction: Mole concept is the method used to define the amount of a compound and the value of 1 mole is $6.022\times {10}^{23}$ Avogadro number.

Answer to Problem 60A

The moles and mass of ${\text{CaCl}}_2$ are $0.025\text{ moles}$ and $2.75\text{ g}$ .

Explanation of Solution

Given, Molar mass of ${\text{CaCl}}_2$ is $110\text{ g/mol}$ .

Volume of ${\text{CaCl}}_2$ is $250\text{ mL}$ .

Molarity of ${\text{CaCl}}_2$ is $0.10\text{ M}$ .

Convert $\text{mL}$ to $\text{L}$ .

  $\begin{array}{c}250\text{mL}=250\times \frac{1\text{L}}{1000\text{mL}}\\ =0.25\text{L}\end{array}$

Substitute, $0.10\text{ M}$ for molarity and $0.25\text{ L}$ for the volume of ${\text{CaCl}}_2$ in the expression for moles of solute.

  $\begin{array}{c}moles{\text{of CaCl}}_2=0.10\text{ M}\times 0.25\text{ L }\\ =0.025\text{ moles}\end{array}$

Substitute, $110\text{ g/mol}$ for molar mass and $0.025\text{ moles}$ for the moles of ${\text{CaCl}}_2$ in the expression for mass of the compound.

  $\begin{array}{c}Mass\text{of Compound}=110\text{ g/mol}\times 0.025moles\\ =2.75\text{ g}\end{array}$

The moles and mass of ${\text{CaCl}}_2$ are $0.025\text{ moles}$ and $2.75\text{ g}$ .

(d)

Interpretation Introduction

Interpretation: The moles and grams of the solute $2.0\text{ L}$ of $0.30\text{ M}$

  ${\text{Na}}_2{\text{SO}}_4$ are to be calculated.

Concept Introduction: Mole concept is the method used to define the amount of a compound and the value of 1 mole is $6.022\times {10}^{23}$ Avogadro number.

Answer to Problem 60A

The moles and mass of ${\text{Na}}_2{\text{SO}}_4$ are $0.6\text{ moles}$ and $85.2\text{ g}$ .

Explanation of Solution

Given, Molar mass of ${\text{Na}}_2{\text{SO}}_4$ is $142\text{ g/mol}$ .

Volume of ${\text{Na}}_2{\text{SO}}_4$ is $2.0\text{ L}$ .

Molarity of ${\text{Na}}_2{\text{SO}}_4$ is $0.30\text{ M}$ .

Substitute, $0.30\text{ M}$ for molarity and $\text{2}\text{.0 L}$ for the volume of ${\text{Na}}_2{\text{SO}}_4$ in the expression for moles of solute.

  $\begin{array}{c}moles{\text{of Na}}_2{\text{SO}}_4=0.30\text{ M}\times 2.0\text{ L}\\ =0.6\text{ moles}\end{array}$

The expression for mass of the compound is as follows:

  $Mass\text{of Compound}=Molar\text{ m}ass\times moles$

Substitute, $142\text{ g/mol}$ for molar mass and $0.6\text{ moles}$ for the moles of ${\text{Na}}_2{\text{SO}}_4$ in the expression for mass of the compound.

  $\begin{array}{c}Mass\text{of Compound}=142\text{ g/mol}\times 0.6moles\\ =85.2\text{ g}\end{array}$

The moles and mass of ${\text{Na}}_2{\text{SO}}_4$ are $0.6\text{ moles}$ and $85.2\text{ g}$ .