Chapter 4, Problem 4.44QP

(a)

Interpretation Introduction

Interpretation:

The mass (in grams) required to have $\text{0}\text{.100 mol}$ of glucose, ${\text{C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}$ has to be calculated.

Concept Introduction:

Mass:

Mass of the compound is calculated by mole of the compound multiplied with molar mass of the compound.

  $\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$

Explanation of Solution

Given,

$\text{0}\text{.100 mol}$ of glucose, ${\text{C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}$

The molar mass of glucose is $180.156\text{ g/mol}$.

  $\frac{180.156{\text{ g C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}}{{\text{1 mol C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}}$

The mass of glucose required to have $\text{0}\text{.100 mol}$ of glucose is calculated as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{Mole}\times \text{Molar}\text{mass}\\ =0.100{\text{ mol C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}\times \frac{180.156{\text{ g C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}}{{\text{1 mol C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}}\\ =18.0156\\ \approx 18.02{\text{ g C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}\end{array}$

Therefore, the mass of ${\text{C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}$ required is $18.02{\text{ g C}}_{\text{6}}{\text{H}}_{12}{\text{O}}_{\text{6}}$.

(b)

Interpretation Introduction

Interpretation:

The mass (in grams) required to have $\text{0}\text{.100 mol}$ of $\text{NaCl}$ has to be calculated.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given,

$\text{0}\text{.100 mol}$ of $\text{NaCl}$

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

  $\frac{58.44\text{ g NaCl}}{\text{1 mol NaCl}}$

The mass of $\text{NaCl}$ required to have $\text{0}\text{.100 mol}$ of $\text{NaCl}$ is calculated as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{Mole}\times \text{Molar}\text{mass}\\ =0.100\text{ mol NaCl}\times \frac{58.44\text{ g NaCl}}{\text{1 mol NaCl}}\\ =5.844\\ \approx 5.8\text{ g NaCl}\end{array}$

Therefore, the mass of $\text{NaCl}$ required is $5.8\text{ g NaCl}$.

(c)

Interpretation Introduction

Interpretation:

The mass (in grams) required to have $\text{0}\text{.100 mol}$ of ethanol, ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ has to be calculated.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given,

$\text{0}\text{.100 mol}$ of ethanol, ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$

The molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is $46.07\text{ g/mol}$.

  $\frac{\text{46}{\text{.07 g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{{\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}$

The mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ required to have $\text{0}\text{.100 mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is calculated as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{Mole}\times \text{Molar}\text{mass}\\ =0.100{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\times \frac{\text{46}{\text{.07 g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{{\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\\ =4.607\\ \approx 4.61{\text{ g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\end{array}$

Therefore, the mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ required is $4.61{\text{ g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$.

(d)

Interpretation Introduction

Interpretation:

The mass (in grams) required to have $\text{0}\text{.100 mol}$ of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Refer part (a)

Explanation of Solution

Given,

$\text{0}\text{.100 mol}$ of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$

The molar mass of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ is $310.18\text{ g/mol}$.

  $\frac{310.18{\text{ g Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}{{\text{1 mol Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}$

The mass of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ required to have $\text{0}\text{.100 mol}$ of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ is calculated as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{Mole}\times \text{Molar}\text{mass}\\ =0.100{\text{ mol Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\times \frac{310.18{\text{ g Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}{{\text{1 mol Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}\\ =31.018\\ \approx 31.02{\text{ g Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\end{array}$

Therefore, the mass of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ required is $31.02{\text{ g Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$.