Mercury is an extremely toxic substance. Inhalation of the vapor is just as dangerous as swallowing the liquid. How many milliliters of mercury will saturate a room that is \(15 \times 12 \times 8.0 \mathrm{ft}\) with mercury vapor at \(25^{\circ} \mathrm{C}\) ? The vapor pressure of \(\mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\) is \(0.00163 \mathrm{~mm} \mathrm{Hg}\) and its density is \(13 \mathrm{~g} / \mathrm{mL}\).

We are given the following information: - Room dimensions: \(15 \times 12 \times 8.0 \mathrm{ft}\) - Temperature: \(25^{\circ} \mathrm{C}\) - Vapor pressure of mercury: \(0.00163 \mathrm{~mm} \mathrm{Hg}\) - Density of mercury: \(13 \mathrm{~g} / \mathrm{mL}\)

To work with SI units, we need to convert the room dimensions from feet to meters using the conversion factor 1 ft = 0.3048 m. So, $$15 \times 12 \times 8.0 \mathrm{ft} \times \left(\frac{0.3048 \mathrm{m}}{1 \mathrm{~ft}}\right)^3 = 68.49264 \mathrm{m^3}$$

We need to find the amount of mercury that corresponds to the given vapor pressure of 0.00163 mm Hg. We will use the Ideal Gas Law formula, \(PV=nRT\), where: - P is the pressure (in \(\mathrm{Pa}\)) - V is the volume (in \(\mathrm{m^3}\)) - n is the moles of mercury - R is the gas constant, \(8.314 \mathrm{J} \cdot \mathrm{K^{-1}} \cdot \mathrm{mol^{-1}}\) - T is the temperature in Kelvin, \(25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{K}\) First, we need to convert the pressure from mm Hg to Pa. $$0.00163 \mathrm{~mm} \mathrm{Hg} \times \frac{101325 \mathrm{Pa}}{760 \mathrm{~mm} \mathrm{Hg}} = 220.9956 \mathrm{~Pa}$$ Then, substitute the given values into the Ideal Gas Law formula to find the moles of mercury (n). $$220.9956 \mathrm{~Pa} \times 68.49264 \mathrm{m^3} = n \times 8.314 \mathrm{J} \cdot \mathrm{K^{-1}} \cdot \mathrm{mol^{-1}} \times 298.15 \mathrm{K}$$ Solving for n, we get $$n = \frac{220.9956 \mathrm{~Pa} \times 68.49264 \mathrm{m^3}}{8.314 \mathrm{J} \cdot \mathrm{K^{-1}} \cdot \mathrm{mol^{-1}} \times 298.15 \mathrm{K}} = 0.5486 \mathrm{~mol}$$

Now that we have the moles of mercury, we can calculate the mass of the mercury using the molar mass of mercury, which is 200.59 \(\mathrm{g/mol}\). $$\text{Mass of Hg}= 0.5486 \mathrm{~mol} \times 200.59 \frac{\mathrm{g}}{\mathrm{mol}} = 110.066 \mathrm{~g}$$

Finally, we can find the volume of mercury using the given density of mercury and the mass calculated in step 4. $$\text{Volume of Hg} = \frac{110.066 \mathrm{~g}}{13 \frac{\mathrm{g}}{\mathrm{mL}}} = 8.4666 \mathrm{~mL}$$ Therefore, 8.4666 mL of mercury will saturate the room with mercury vapor at \(25^{\circ} \mathrm{C}\).