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Chapter 13: Problem 37

The Environmental Protection Agency has a limit of 15 ppm for the amount of lead in drinking water. If a \(1.000 \mathrm{mL}\) sample of water at \(20^{\circ} \mathrm{C}\) contains \(15 \mathrm{ppm}\) of lead, how many lead ions are there in this sample of water? What is the mole fraction of lead ion in solution?

Short Answer

Expert verified
There are approximately \(2.91 \times 10^{18}\) lead ions in a water sample of 1.000 ml with 15 ppm of lead, and the mole fraction of lead ion in the solution is \(8.7 \times 10^{-8}\).

Step by step solution

01

Convert ppm to moles

ppm means parts per million. In this case, it says there are 15 parts of lead in 1 million parts of water. Since there are 1000 mL in one liter, there is 1 mg of lead in this sample. Since the molar mass of lead is approximately 207 g/mol, to convert mg to g, divide by 1000. So there are approximately \(\frac{1}{1000} \times \(\frac{1}{207} = 4.83 \times 10^{-6}\) mol of lead.
02

Compute the number of ions

Using Avogadro's number, which states that one mole of a substance contains \(6.022 \times 10^{23}\) particles, the number of lead ions can be computed as \(4.83 \times 10^{-6}\) mol \(\times 6.022\times10^{23}\) mol\(^{-1}\) = \(2.91 \times 10^{18}\) ions.
03

Calculate mole fraction

The mole fraction of a substance is the number of moles of the substance divided by the total number of moles in the solution. Here, the total moles of the solution need to be calculated first. Given the density of water to be 1g/mL at \(20^{\circ} \mathrm{C}\), then the mass of one liter of water is 1000g. Thus, its moles would be \(\frac{1000}{18.02} = 55.5\) moles (molar mass of water is 18.02). Hence, the mole fraction of lead would be \(\frac{4.83 \times 10^{-6}}{55.5} = 8.7 \times 10^{-8}\).

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Most popular questions from this chapter

Calculate the mole fraction of solute in the following aqueous solutions: (a) \(21.7 \% \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) by mass; (b) \(0.684 m \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\) (urea).

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A water sample is found to have 9.4 ppb of chloroform, \(\mathrm{CHCl}_{3} .\) How many grams of \(\mathrm{CHCl}_{3}\) would be found in a glassful \((250 \mathrm{mL})\) of this water?

Calculate the vapor pressure at \(20^{\circ} \mathrm{C}\) of a saturated solution of the nonvolatile solute, urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2},\) in methanol, \(\mathrm{CH}_{3} \mathrm{OH} .\) The solubility is \(17 \mathrm{g}\) urea/100 \(\mathrm{mL}\) methanol. The density of methanol is \(0.792 \mathrm{g} / \mathrm{mL}\), and its vapor pressure at \(20^{\circ} \mathrm{C}\) is \(95.7 \mathrm{mmHg}\).

Nitrobenzene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2},\) and benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) are completely miscible in each other. Other properties of the two liquids are nitrobenzene: \(\mathrm{fp}=5.7^{\circ} \mathrm{C}, K_{\mathrm{f}}=\) \(8.1^{\circ} \mathrm{C} m^{-1} ;\) benzene: \(\mathrm{fp}=5.5^{\circ} \mathrm{C}, K_{\mathrm{f}}=5.12^{\circ} \mathrm{C} m^{-1} . \mathrm{It}\) is possible to prepare two different solutions with these two liquids having a freezing point of \(0.0^{\circ} \mathrm{C}\) What are the compositions of these two solutions, expressed as mass percent nitrobenzene?
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