91Ó°ÊÓ

The mass of a substance made up of an equal number of fundamental units is defined as a mole.

a)

Multiply the molar mass of Chromium (III) sulfate decahydrate by the stated number of moles.

The molar massof$C{r}_2{\left(S{O}_4\right)}_3×10H_{2}O$ is $572.41\frac{g}{mol}$.

$\begin{array}{c}Mass\text{of}C{r}_2{\left(S{O}_4\right)}_3â‹\dots 10H_{2}O=8.42\text{mol}C{r}_2{\left(S{O}_4\right)}_3â‹\dots 10H_{2}O×\left(\frac{572.41\text{g}C{r}_2{\left(S{O}_4\right)}_3â‹\dots 10H_{2}O}{\text{mol}C{r}_2{\left(S{O}_4\right)}_3â‹\dots 10H_{2}O}\right)\\ =\boxed{4819.69\text{g}C{r}_2{\left(S{O}_4\right)}_3â‹\dots 10H_{2}O}\end{array}$

b)

Multiplythe number of molecules of dichlorineheptaoxide, $C{l}_2{O}_7$, by the reciprocal of the number of molecules.

The Avogadro's number and the molar mass of ${\mathrm{Cl}}_2{O}_7$, respectively, are .$182.90\frac{g}{mol}$

$\begin{array}{c}Mass\text{of}C{l}_2{O}_7=1.83×{10}^{24}\text{molecules}C{l}_2{O}_7×\left(\frac{molC{l}_2{O}_7}{6.022×{10}^{23}\text{molecules}C{l}_2{O}_7}\right)×\left(\frac{182.90\text{g}C{l}_2{O}_7}{molC{l}_2{O}_7}\right)\\ =\boxed{555.81\text{g}C{l}_2{O}_7}\end{array}$

c)

We multiply the given number of moles in 6.2 g of lithium sulfate, ${\text{Li}}_2{\text{SO}}_4$, by the reciprocal of its molar mass,$109.95\frac{g}{mol}$. $\begin{array}{c}Molesof{\text{Li}}_2{\text{SO}}_4=6.2{\text{gLi}}_2{\text{SO}}_4×\left(\frac{mol{\text{Li}}_2{\text{SO}}_4}{109.95{\text{gLi}}_2{\text{SO}}_4}\right)\\ =5.64×{10}^{−2}Molof{\text{Li}}_2{\text{SO}}_4\end{array}$

We multiply the calculated number of moles of by ${\text{Li}}_2{\text{SO}}_4$the Avogadro's number to get the formula units (FU).

$\begin{array}{c}Formula\text{units}of{\text{Li}}_2{\text{SO}}_4=5.64{\text{molLi}}_2{\text{SO}}_4×\left(\frac{6.022×{10}^{23}FU{\text{Li}}_2{\text{SO}}_4}{mol{\text{NLi}}_2{\text{SO}}_4}\right)\\ =\boxed{3.40×{10}^{22}FU{\text{Li}}_2{\text{SO}}_4}\end{array}$

d)

One mole of ${\mathrm{Li}}_{2}S{O}_4$contains two moles of ions ${\text{Li}}^{\text{+}}$. In section (c), the number of moles of ions in the mass of the compound isas follows.

$\begin{array}{c}MolesofL{i}^{+}\text{atoms}=5.64×{10}^{−2}\text{mol}L{i}_{2}S{O}_4×\left(\frac{2molL{i}^{+}ions}{molL{i}_{2}S{O}_4}\right)\\ =0.113molL{i}^{+}\text{ions}\end{array}$

Thenumber of moles of role="math" localid="1656649256211" ${\text{Li}}^{\text{+}}$ions is multiplied by Avogadro's number.

$\begin{array}{c}MolesofL{i}^{+}\text{atoms}=0.113\text{mol}L{i}^{+}\text{ions}×\left(\frac{6.022×{10}^{23}L{i}^{+}ions}{molL{i}^{+}\text{ions}}\right)\\ =6.80×{10}^{22}L{i}^{+}\text{ions}\end{array}$

Because one mole of$S{O}_4^{2−}$ions and one mole of S atoms make up one mole of ${\text{Li}}_2{\text{SO}}_4$, their mole numbers are numerically equal. As a result, the amount of$S{O}_4^{2−}$ions and S atoms in ${\text{Li}}_2{\text{SO}}_4$equals the number of formula units.

$\begin{array}{l}No.{\text{ofSO}}_4^{2−}\text{ions}=3.40×{10}^{22}{\text{SO}}_4^{2−}\text{ions}\\ No.\text{ofSatoms}=3.40×{10}^{22}\text{Satoms}\end{array}$

One mole of${\text{Li}}_2{\text{SO}}_4$contains four moles of O atoms. The following is the number of moles of O atoms in the mass of the compound in section(c).

$\begin{array}{c}MolesofO\text{atoms}=5.64×{10}^{−2}{\text{molLi}}_2{\text{SO}}_4×\left(\frac{4mol\text{O}atoms}{mol{\text{Li}}_2{\text{SO}}_4}\right)\\ =0.226mol\text{Oatoms}\end{array}$

The number of moles of O atoms is multiplied by Avogadro's number.

$\begin{array}{c}No.ofO\text{atoms}=0.226\text{mol}O\text{atoms}×\left(\frac{6.022×{10}^{23}\text{O}atoms}{mol\text{Oatoms}}\right)\\ =\boxed{1.36×{10}^{23}\text{Oatoms}}\end{array}$