Chapter 11, Problem 7PEB

(a)

To determine

The percent of average amount of lead in a person.

Answer to Problem 7PEB

Solution:

$2.5\times {10}^{-5}\%$

Explanation of Solution

Given:

The blood of the average American contained $0.25\text{ ppm}$ lead.

The danger level of lead poisoning is $0.80\text{ ppm}$.

Formula used:

Write the expression for the percent of concentration.

$\%\text{ concentartion}=\frac{\text{ppm of lead}}{1\times {10}^4\frac{\text{ppm}}{\%}}$

Explanation:

Recall the expression for the percent of concentration.

$\%\text{ concentartion}=\frac{\text{ppm of lead}}{1\times {10}^4\frac{\text{ppm}}{\%}}$

Substitute $0.25\text{ ppm}$ for ppm of lead.

$\begin{array}{c}\%\text{ concentartion}=\frac{0.25\text{ ppm}}{1\times {10}^4\frac{\text{ppm}}{\%}}\\ =2.5\times {10}^{-5}\%\end{array}$

Conclusion:

The percent of the average person lead are $2.5\times {10}^{-5}\%$.

(b)

To determine

The lead would be in an average of $80\text{ kg}$ person.

Answer to Problem 7PEB

Solution:

$0.02\text{ g}$

Explanation of Solution

Given:

The blood of the average American contained $0.25\text{ ppm}$ lead.

The danger level of lead poisoning is $0.80\text{ ppm}$.

Formula used:

Write the expression for the percent of part.

$\%\text{ of part}=\frac{\text{mass of part}}{\text{mass of whole}}\times 100\%\text{ of whole}$

Explanation:

Convert kg to g.

$\begin{array}{c}80\text{ kg}=80\text{ kg}\left(\frac{{10}^3\text{ g}}{1\text{ kg}}\right)\\ =80,000\text{ g}\end{array}$

Refer the part (a) to the value of $\%\text{ of part}$.

$2.5\times {10}^{-5}\%$

Recall the expression for the percent of part.

$\%\text{ of part}=\frac{\text{mass of part}}{\text{mass of whole}}\times 100\%\text{ of whole}$

Rearrange:

$\text{mass of part}=\frac{\text{mass of whole}}{100\%\text{ of whole}}\times \%\text{ of part}$

Substitute $2.5\times {10}^{-5}\%$ for $\%\text{ of part}$ and $80,000\text{ g}$ for mass of whole.

$\begin{array}{c}\text{mass of part}=\frac{80,000\text{ g of whole}}{100\%\text{ of whole}}\times 2.5\times {10}^{-5}\%\\ =0.02\text{ g}\end{array}$

Conclusion:

The lead would be in an average $80\text{ kg}$ person are $0.02\text{ g}$.

(c)

To determine

The lead would be in an average person need to accumulate to reach the danger level.

Answer to Problem 7PEB

Solution:

$8\times {10}^{-5}\%$

Explanation of Solution

Given:

The blood of the average American contained $0.25\text{ ppm}$ lead.

The danger level of lead poisoning is $0.80\text{ ppm}$.

Formula used:

Write the expression for the percent of concentration.

$\%\text{ concentartion}=\frac{\text{ppm of lead}}{1\times {10}^4\frac{\text{ppm}}{\%}}$

Explanation:

Recall the expression for the percent of concentration.

$\%\text{ concentartion}=\frac{\text{ppm of lead}}{1\times {10}^4\frac{\text{ppm}}{\%}}$

Substitute $0.80\text{ ppm}$ for ppm of lead.

$\begin{array}{c}\%\text{ concentartion}=\frac{0.80\text{ ppm}}{1\times {10}^4\frac{\text{ppm}}{\%}}\\ =8\times {10}^{-5}\%\end{array}$

Conclusion:

The lead would be in an average person need to accumulate to reach the danger level

are $8\times {10}^{-5}\%$.