Trace organic compounds in the atmosphere are first concentrated and then measured by gas chromatography. In the concentration step, several liters of air are pumped through a tube containing a porous substance that traps organic compounds. The tube is then connected to a gas chromatograph and heated to release the trapped compounds. The organic compounds are separated in the column and the amounts are measured. In an analysis for benzene and toluene in air, a \(3.00-\mathrm{L}\) sample of air at 748 torr and \(23^{\circ} \mathrm{C}\) was passed through the trap. The gas chromatography analysis showed that this air sample contained \(89.6\) ng benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) and \(153 \mathrm{ng}\) toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right) .\) Calculate the mixing ratio (see Exercise 121 ) and number of molecules per cubic centimeter for both benzene and toluene.

First, we need to convert the given information into appropriate units. - Air sample volume: \(3.00 L = 3.00 \times 10^{-3} m^3\) - Pressure: \(748 torr = 748 \frac{101.325 kPa}{760 torr} = 100.5 kPa\) - Temperature: \(23^{\circ}C = 23 + 273.15 = 296.15 K\) - Benzene mass: \(89.6 ng = 89.6 \times 10^{-9} g\) - Toluene mass: \(153 ng = 153 \times 10^{-9} g\)

Next, we need to calculate the moles of benzene and toluene in the air sample. Molecular weight of benzene (\(C_6H_6\)) = \(12\times6 + 1\times6 = 78\) g/mol Molecular weight of toluene (\(C_7H_8\)) = \(12\times7 + 1\times8 = 92\) g/mol Moles of benzene = \(\frac{89.6\times10^{-9}\;g}{78\;g/mol} = 1.15\times10^{-10} \;mol\) Moles of toluene = \(\frac{153\times10^{-9}\;g}{92\;g/mol} = 1.66\times10^{-10} \;mol\)

Now, we need to find out the moles of air in the sample using the ideal gas law equation, \(PV = nRT\): \(n = \frac{PV}{RT}\) where: P = pressure of air sample = 100.5 kPa V = volume of air sample = \(3.00\times10^{-3}\) m³ \(R = 8.314\;\frac{J}{mol\cdot K}\) (universal gas constant) T = temperature of air sample = 296.15 K Moles of air = \(\frac{(100.5\:kPa)(3.00\times10^{-3}\;m^3)}{(8.314\;\frac{J}{mol\cdot K})(296.15 K)} = 0.121\;mol\)

The mixing ratio is defined as the ratio of the moles of a substance present to the moles of air in the sample. Mixing ratio of benzene = \(\frac{1.15\times10^{-10}\;mol}{0.121\;mol} = 9.50\times10^{-10}\) Mixing ratio of toluene = \(\frac{1.66\times10^{-10}\;mol}{0.121\;mol} = 1.37\times10^{-9}\)

First, let's calculate the concentration of benzene and toluene in the air sample (moles per cubic meter): Benzene concentration = \(\frac{1.15\times10^{-10}\;mol}{3.00\times10^{-3}\;m^3} = 3.83\times10^{-8}\;\frac{mol}{m^3}\) Toluene concentration = \(\frac{1.66\times10^{-10}\;mol}{3.00\times10^{-3}\;m^3} = 5.53\times10^{-8}\;\frac{mol}{m^3}\) Now, we can calculate the number of molecules per cubic centimeter using Avogadro's number (\(N_A = 6.022\times10^{23}\; \frac{molecules}{mol}\)): Number of benzene molecules per \(cm^3\) = \(3.83\times10^{-8}\;\frac{mol}{m^3} \times 6.022\times10^{23}\;\frac{molecules}{mol} \times \frac{1\:m^3}{10^6\:cm^3} = 2.30\times10^{10}\;\frac{molecules}{cm^3}\) Number of toluene molecules per \(cm^3\) = \(5.53\times10^{-8}\;\frac{mol}{m^3} \times 6.022\times10^{23}\;\frac{molecules}{mol} \times \frac{1\:m^3}{10^6\:cm^3} = 3.33\times10^{10}\;\frac{molecules}{cm^3}\) So, the mixing ratio for benzene is \(9.50\times10^{-10}\), and there are \(2.30\times10^{10}\) molecules of benzene per cubic centimeter. For toluene, the mixing ratio is \(1.37\times10^{-9}\), and there are \(3.33\times10^{10}\) molecules per cubic centimeter.