Q3.28P - Chemistry: Molecular Nature Of Matter And Change

Question

The mineral galena is composed of lead(II) sulfide and has an average density of $\mathbf{7}\mathbf{.46}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$.

1. How many moles of lead(II) sulfide are in $\mathbf{1}\mathbf{.00}{\mathbf{ft}}^{\mathbf{3}}$ of galena?
2. How many lead atoms are in $\mathbf{1}\mathbf{.00}{\mathbf{dm}}^{\mathbf{3}}$ of galena?

Step-by-Step Solution

Answer

1. The Moles of lead(II) sulfide are in 1.00 ft³ of galena is 882.9 mol.
2. The lead atoms are in 1.00 dm³ of galena is $\mathbf{1}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{25}}\mathbf{atoms}$.

Step 1: To Calculate how many moles of lead(II) sulfide are in 1.00 ft 3 of galena,

The V(galena) are in $1{\mathrm{ft}}^3=28316.85{\mathrm{cm}}^3$

The density of d(galena) is $7.46\mathrm{g}/{\mathrm{cm}}^3$

The moles of lead(II) sulfide m(galena)$=\mathrm{d}\times \mathrm{V}$

$=28316.85{\mathrm{cm}}^3\times 7.46\mathrm{g}/{\mathrm{cm}}^3$

The moles of lead(II) sulfide m(galena)$=211243.701\mathrm{g}$

$\begin{array}{c}\mathrm{Mr}\left(\mathrm{PbS}\right)=\mathrm{Ar}\left(\mathrm{Pb}\right)\times \mathrm{Ar}\left(\mathrm{S}\right)\\ \mathrm{Mr}\left(\mathrm{PbS}\right)=207.20+32.06\\ \mathrm{Mr}\left(\mathrm{PbS}\right)=239.26\mathrm{g}/\mathrm{mol}\\ \mathrm{n}\left(\mathrm{PbS}\right)=\mathrm{m}/\mathrm{Mr}=\frac{211243.7\mathrm{g}}{239.26\mathrm{g}/\mathrm{mol}}\\ \mathrm{n}\left(\mathrm{PbS}\right)=882.9\mathrm{mol}\end{array}$

The Moles of lead(II) sulfide are in 1.00 ft³ of galena is 882.9 mol.

Step 2: How many lead atoms are in 1.00 dm 3 of galena,

The V=$1{\mathrm{dm}}^3={10}^3{\mathrm{cm}}^3$

$\mathrm{m}=\mathrm{d}\times \mathrm{V}=7.46\mathrm{g}/{\mathrm{cm}}^3\times {10}^3{\mathrm{cm}}^3$

m(galena)= 7460gm(galena)=239.26 g/cm³

n(galena) =$\mathrm{m}/\mathrm{Mr}=\mathrm{n}\left(\mathrm{Pb}\right)$

$\mathrm{n}\left(\mathrm{galena}\right)=\frac{7460\mathrm{g}}{239.26\mathrm{g}/\mathrm{mol}}$.n(galena)= 31.18 mol.

$\begin{array}{c}\mathrm{n}\left(\mathrm{Pb}\right)=\mathrm{n}\left(\mathrm{galena}\right)=31.18\mathrm{mol}\\ \mathrm{n}\left(\mathrm{Pb}\right)=\mathrm{n}\left(\mathrm{Pb}\right)\times {\mathrm{N}}_{\mathrm{A}}\\ \mathrm{n}\left(\mathrm{Pb}\right)=31.18\times 6.022\times 10{}^{23}\\ \mathrm{N}\left(\mathrm{Pb}\right)=1.88\times 10{}^{25}\end{array}$

The lead atoms are in 1.00 dm³ of galena is $1.88\times {10}^{25}$ atoms.