Chapter U4.81, Problem 6E

a.

Interpretation Introduction

Interpretation: The number of moles of sodium cation present in 1.0 L of $\text{0}\text{.10 M NaCl}$ needs to be determined.

Concept Introduction: The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ - (1)

Answer to Problem 6E

The number of moles of sodium cation present in 1.0 L of $\text{0}\text{.10 M NaCl}$ is $\text{0}{\text{.10 mol Na}}^{+}$ .

Explanation of Solution

The equation (1) is rearranged as:

  $\text{number of moles of solute = Molarity}\times \text{Volume of solution (L)}$

The given solution is 1.0 L of $\text{0}\text{.10 M NaCl}$ so, the number of moles in the given solution is calculated as:

  $\begin{array}{l}\text{number of moles of solute = 0}\text{.1 mol/L}\times \text{1}\text{.0 L}\\ \text{number of moles of solute = 0}\text{.1 mol}\end{array}$

So, the number of moles $\text{NaCl}$ in the given solution is $\text{0}\text{.10 mol}$ .

Since, one mole of $\text{NaCl}$ will give one mole of ${\text{Na}}^{+}$ so, $\text{0}\text{.10 mol NaCl}$ will give $\text{0}{\text{.10 mol Na}}^{+}$ .

Hence, the number of moles of sodium cation present in 1.0 L of $\text{0}\text{.10 M NaCl}$ is $\text{0}{\text{.10 mol Na}}^{+}$ .

b.

Interpretation Introduction

Interpretation: The number of moles of sodium cation present in 1.0 L of $\text{3}{\text{.0 M Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ needs to be determined.

Concept Introduction: The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ - (1)

Answer to Problem 6E

The number of moles of sodium cation present in 1.0 L of $\text{3}{\text{.0 M Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{6}{\text{.0 mol Na}}^{+}$ .

Explanation of Solution

The equation (1) is rearranged as:

  $\text{number of moles of solute = Molarity}\times \text{Volume of solution (L)}$

The given solution is 1.0 L of $\text{3}{\text{.0 M Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ so, the number of moles in the given solution is calculated as:

  $\begin{array}{l}\text{number of moles of solute = 3}\text{.0 mol/L}\times \text{1}\text{.0 L}\\ \text{number of moles of solute = 3}\text{.0 mol}\end{array}$

So, the number of moles ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the given solution is $\text{3}\text{.0 mol}$ .

Since, one mole of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ will give two moles of ${\text{Na}}^{+}$ so, $\text{3}{\text{.0 mol Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ will give $\text{3}\times {\text{2 = 6 mol Na}}^{+}$ .

Hence, the number of moles of sodium cation present in 1.0 L of $\text{3}{\text{.0 M Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{6}{\text{.0 mol Na}}^{+}$ .

c.

Interpretation Introduction

Interpretation: The number of moles of sodium cation present in 1.0 L of $\text{0}{\text{.30 M Na}}_3{\text{PO}}_{\text{4}}$ needs to be determined.

Concept Introduction: The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ - (1)

Answer to Problem 6E

The number of moles of sodium cation present in 1.0 L of $\text{0}{\text{.30 M Na}}_3{\text{PO}}_{\text{4}}$ is $\text{0}{\text{.9 mol Na}}^{+}$ .

Explanation of Solution

The equation (1) is rearranged as:

  $\text{number of moles of solute = Molarity}\times \text{Volume of solution (L)}$

The given solution is 1.0 L of $\text{0}{\text{.30 M Na}}_3{\text{PO}}_{\text{4}}$ so, the number of moles in the given solution is calculated as:

  $\begin{array}{l}\text{number of moles of solute = 0}\text{.3 mol/L}\times \text{1}\text{.0 L}\\ \text{number of moles of solute = 0}\text{.3 mol}\end{array}$

So, the number of moles ${\text{Na}}_3{\text{PO}}_{\text{4}}$ in the given solution is $\text{0}\text{.3 mol}$ .

Since, one mole of ${\text{Na}}_3{\text{PO}}_{\text{4}}$ will give three moles of ${\text{Na}}^{+}$ so, $\text{0}{\text{.3 mol Na}}_3{\text{PO}}_{\text{4}}$ will give $\text{0}\text{.3}\times 3\text{ = 0}{\text{.9 mol Na}}^{+}$ .

Hence, the number of moles of sodium cation present in 1.0 L of $\text{0}{\text{.30 M Na}}_3{\text{PO}}_{\text{4}}$ is $\text{0}{\text{.9 mol Na}}^{+}$ .