Chapter 3, Problem 3.121QP

Calculate the number of cations and anions in each of the following compounds: (a) 0.764 g of CsI, (b) 72.8 g of K₂Cr₂O₇, (c) 6.54 g of Hg₂(NO₃)₂.

Interpretation Introduction

Interpretation:

From the given mass of $\text{CsI}$, the number of cations and anions in $\text{CsI}$ has to be calculated.

Concept Introduction:

Molar mass:

• Molar mass of a molecule can be calculated from its molecular formula by taking the sum of atomic masses of all the elements present in it.
• Conversion formula for mass of a molecule into number moles,

$Numberofmoles=\frac{Massingrams}{Molarmass}$

Answer to Problem 3.121QP

The number of iodide ions is $\text{1}\text{.77}{\text{×10}}^{\text{21}}{\text{I}}^{-}\text{ions}$

The number of cesium ions is $\text{1}\text{.77}{\text{×10}}^{\text{21}}{\text{I}}^{-}\text{ions}$

Explanation of Solution

Given,

Mass of $\text{CsI}$ is $\text{0}\text{.764g}$

The number of cesium ions and iodide ions are calculated using molar mass of $\text{CsI}$ and Avogadro number as,

$\text{0}\text{.764 g CsI}\text{×}\frac{\text{1 mol CsI}}{\text{259}\text{.8 g CsI}}\text{×}\frac{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{CsI}}{\text{1}\text{mol}\text{CsI}}\text{×}\frac{{\text{1Cs}}^{\text{+}}\text{ion}}{\text{1CsI}}\text{=}\text{1}\text{.77}{\text{×10}}^{\text{21}}{\text{Cs}}^{\text{+}}\text{ions}$

The ratio of cesium ion and iodide ion is $1:1$.  Hence, the number of iodide ions $\text{1}\text{.77}{\text{×10}}^{\text{21}}{\text{I}}^{-}\text{ions}$ and number of cesium ions is $\text{1}\text{.77}{\text{×10}}^{\text{21}}{\text{I}}^{-}\text{ions}$

Interpretation Introduction

Interpretation:

From the given mass of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$, the number of cations and anions in ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ has to be calculated.

Concept Introduction:

Molar mass:

• Molar mass of a molecule can be calculated from its molecular formula by taking the sum of atomic masses of all the elements present in it.
• Conversion formula for mass of a molecule into number moles,

$Numberofmoles=\frac{Massingrams}{Molarmass}$ 

Answer to Problem 3.121QP

The number of chromate ions is $1.49{\text{×10}}^{\text{23}}{\text{Cr}}_{\text{2}}{\text{O}}_7{}^{2-}\text{ions}$

The number of potassium ion is $\text{2}\text{.98}{\text{×10}}^{\text{23}}{\text{K}}^{\text{+}}\text{ions}$

Explanation of Solution

Given,

Mass of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is $\text{72}\text{.8g}$

The number of potassium ions and chromate ions are calculated using molar mass of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ and Avogadro number as,

$\text{72}{\text{.8 g K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{×}\frac{{\text{1 mol K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{294}{\text{.2 g K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}\text{×}\frac{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{1}\text{mol}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}\text{×}\frac{{\text{2K}}^{\text{+}}\text{ion}}{{\text{1K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}\text{=}\text{2}\text{.98}{\text{×10}}^{\text{23}}{\text{K}}^{\text{+}}\text{ions}$

The ratio of potassium ion and chromate ion is $2:1$ .  Hence, the number of chromate ions is $1.49{\text{×10}}^{\text{23}}{\text{Cr}}_{\text{2}}{\text{O}}_7{}^{2-}\text{ions}$ and number of potassium ion is $\text{2}\text{.98}{\text{×10}}^{\text{23}}{\text{K}}^{\text{+}}\text{ions}$

Interpretation Introduction

Interpretation:

From the given mass of ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$, the number of cations and anions in ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ have to be calculated.

Concept Introduction:

Molar mass:

• Molar mass of a molecule can be calculated from its molecular formula by taking the sum of atomic masses of all the elements present in it.
• Conversion formula for mass of a molecule into number moles,

$Numberofmoles=\frac{Massingrams}{Molarmass}$ 

Answer to Problem 3.121QP

The number of nitrate ions is $1.50{\text{×10}}^{\text{22}}{\text{NO}}_3{}^{-}\text{ions}$

The number of mercury ion is $7.50{\text{×10}}^{\text{21}}{\text{Hg}}_{\text{2}}{}^{\text{2+}}\text{ions}$

Explanation of Solution

Given,

Mass of ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ is $\text{72}\text{.8g}$

The number of mercury ions and nitrate ions are calculated using molar mass of ${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ and Avogadro number as,

$\text{6}{\text{.54g Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}\text{×}\frac{{\text{1 mol Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}}{\text{294}{\text{.2 g Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}}\text{×}\frac{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}{\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}}{\text{1}\text{mol}{\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}}\text{×}\frac{{\text{1Hg}}_{\text{2}}{}^{\text{2+}}\text{ion}}{{\text{1Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}}\text{=}7.50{\text{×10}}^{\text{21}}{\text{Hg}}_{\text{2}}{}^{\text{2+}}\text{ions}$

The ratio of mercury ions and nitrate ions is $1:2$ .  Hence, the number of nitrate ions is $1.50{\text{×10}}^{\text{22}}{\text{NO}}_3{}^{-}\text{ions}$ and number of mercury ion is $7.50{\text{×10}}^{\text{21}}{\text{Hg}}_{\text{2}}{}^{\text{2+}}\text{ions}$