Respuesta :
If the maximum level that the EPA considers safe for lead air pollution is 1.5 μg/m³ and the total lung volume is 5.60 L, then 2.44 ×10¹³ atoms of lead will be in the lungs.
FURTHER EXPLANATION
The solution to this problem involves multistep dimensional analysis using the conversion factors below:
- 1.5 μg lead = 1 m³ of air
- 1 m³ = 1000 L
- 1 mol of lead = 207.2 g of lead
- 1 g = 10⁶ μg
- 1 mol lead = 6.022×10²³ lead atoms
1. To solve the problem, first convert the volume of the lungs to m³.
[tex]volume \ of \ lungs \ = 5.60 \ L \times \frac{1 \ m^3}{1000 \ L}\\\\\boxed {volume\ of \ lungs \ = 5.6 \times 10^{-3} \ m^3}[/tex]
2. Determine how many lead atoms are present if the whole volume of the lungs is filled with air.
[tex]mass \ of \ lead \ inhaled \ = 5.6 \times 10^{-3} \ m^3 \ \times \frac{1.56 \mu g \ lead}{1 \ m^3 \ air}\\\\\boxed {mass\ of \ lead \ inhaled \ = 8.4 \times 10^{-3} \ \mu g \ of \ lead}[/tex]
3. Convert the mass of lead into moles of lead.
[tex]moles \ of \ lead \ = 8.4 \times 10^{-3} \ \mu g \ lead \times \frac{1 \ g}{10^6 \ \mu g} \times \frac{1 \ mol \ lead}{207.2 \ g \ lead}\\\\\boxed {moles \ of\ lead \ = 4.0541 \times 10^{-11} \ mol \ lead}\\[/tex]
4. Determine how many atoms are in the given moles of lead using Avogadro's number.
[tex]number \ of \ lead \ atoms \ = 4.0541 \times 10^{-11} \ mol \ lead \times \frac{6.022 \times 10^{23} \ atoms}{1 \ mol \ lead}\\\\\boxed {number \ of \ lead \ atoms \ = 2.4414 \times 10^{13} atoms}[/tex]
Since the number of significant figures in the given is 3 (5.60 L), the final answer must also be expressed with 3 significant figures. Therefore,
[tex]\boxed {\boxed {number \ of \ lead \ atoms \ = 2.44 \times 10^{13} \ atoms}}[/tex]
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Keywords: unit conversion, lead
[tex]\boxed{1845.99 \times {{10}^{23}}{\text{ atoms}}}[/tex] of lead are present in the lungs if its volume is 5.60 L.
Further Explanation:
Avogadro’s number gives information about the number of atoms or molecules present in one mole of the substance. The value of Avogadro’s number is [tex]{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}[/tex]. These units can either be atoms or molecules.
Density is defined as the ratio of mass per unit volume. It is an intensive physical property of the substance.
The formula to calculate the density of a substance is as follows:
[tex]{\text{Density}} = \dfrac{{{\text{Mass}}}}{{{\text{Volume}}}}[/tex] ...... (1)
Rearrange equation (1) for mass.
[tex]{\text{Mass}} = \left( {{\text{Density}}} \right)\left( {{\text{Volume}}} \right)[/tex] ....... (2)
The volume is to be converted into [tex]{\text{c}}{{\text{m}}^{\text{3}}}[/tex]. The conversion factor for this is,
[tex]1{\text{ L}} = {10^3}{\text{ c}}{{\text{m}}^3}[/tex]
Therefore the volume can be calculated as follows:
[tex]\begin{aligned}{\text{Volume}}&=\left( {5.60{\text{ L}}} \right)\left( {\frac{{{{10}^3}{\text{ c}}{{\text{m}}^3}}}{{1{\text{ L}}}}} \right)\\&= 5600{\text{ c}}{{\text{m}}^3}\\\end{aligned}[/tex]
Substitute [tex]11.342{\text{ g/c}}{{\text{m}}^3}[/tex] for the density and [tex]5600{\text{ c}}{{\text{m}}^3}[/tex] for the volume in equation (2) to calculate the mass of lead.
[tex]\begin{aligned}{\text{Mass of lead}} &= \left( {\frac{{11.342{\text{ g}}}}{{1{\text{ c}}{{\text{m}}^3}}}} \right)\left( {5600{\text{ c}}{{\text{m}}^3}} \right)\\&= 63515.2{\text{ g}}\\\end{aligned}[/tex]
The formula to calculate the moles of lead is as follows:
[tex]{\text{Moles of lead}} = \dfrac{{{\text{Mass of lead}}}}{{{\text{Molar mass of lead}}}}[/tex] ...... (3)
Substitute 63515.2 g for the mass of lead and 207.2 g/mol for the molar mass of lead in equation (3).
[tex]\begin{aligned}{\text{Moles of lead}} &= \frac{{{\text{63515}}{\text{.2 g}}}}{{{\text{207}}{\text{.2 g/mol}}}}\\&= 306.541{\text{ mol}}\\\end{aligned}[/tex]
Since one mole of lead consists of [tex]{\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{atoms}}[/tex]. Therefore the number of atoms in 306.541 moles of lead can be calculated as follows:
[tex]\begin{aligned}{\text{Atoms of lead}} &= \left( {306.541{\text{ mol}}} \right)\left( {\frac{{6.022 \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{atoms}}}}{{1{\text{ mol}}}}} \right)\\&= 1845.99 \times {10^{23}}{\text{ atoms}}\\\end{aligned}[/tex]
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Answer details:
Grade: Senior School
Chapter: Mole concept
Subject: Chemistry
Keywords: lead, molar mass, moles, atoms, density, mass, volume, conversion factor, 306.541 mol, [tex]1845*10^23[/tex] atoms, Avogadro’s number.
