Question

How much of the water (in mL) contains 140 mg of Pb? (Assume a density of 1.0 g/mL.)

Answer 1

If the contaminated water contains 0.0013% Pb by mass, we can write the following equation to find the volume of Pb:

$$\frac{m_\text{Pb}}{m_\text{water}+m_\text{Pb}}=\omega_\text{Pb}.$$

Therefore:

$$m_\text{water}=\frac{m_\text{Pb}}{\omega_\text{water}}-m_\text{Pb}=m_\text{Pb}\bigg(\frac{1}{\omega_\text{water}}-1\bigg)=\\ \space\\ =145\bigg(\frac{1}{0.0013/100}-1\bigg)=11153700\text{ mg}.$$

$$V_\text{water}=m_\text{water}\rho_\text{water}=11153.7\cdot1=11153.7\text{ mL}.$$

This equals to 11.15 L of water without Pb.