Question

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 \(13 \mathrm{~g} / \mathrm{mL}\).

Short answer

Answer: Approximately 0.419 mL of mercury is needed to saturate the room.

Step-by-step solution

Convert the room dimensions to L

First, we need to convert the room dimensions from feet to liters. We can use the following conversion factors: - 1 foot = 0.3048 meters - 1 meter = 1000 liters The volume of the room in liters is: \[((15 \times 12 \times 8.0) \times 0.3048^3) \times 1000 = 33,173.6 \text{ L}\]

Use the Ideal Gas Law

We will use the ideal gas law to find the number of moles of mercury vapor required to saturate the room. \[\text{PV} = \text{nRT}\] Where: P = pressure (in atm) V = volume (in L) n = number of moles of mercury vapor R = ideal gas constant (0.0821 L atm/(mol K)) T = temperature (in K) First, convert the vapor pressure of mercury from mmHg to atm. 1 atm = 760 mmHg \[0.00163 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 2.14 \times 10^{-6} \text{ atm}\] Next, convert the temperature from Celsius to Kelvin. \[25^{\circ} \text{C} + 273.15 = 298.15 \text{ K}\] Now, we can find the number of moles of mercury vapor. \[n = \frac{PV}{RT} = \frac{(2.14 \times 10^{-6} \text{ atm})(33,173.6 \text{ L})}{(0.0821 \text{ L atm/(mol K)})(298.15 \text{ K})} = 0.0272 \text{ mol }\]

Calculate mass of mercury vapor

Now, we will find the mass of mercury vapor using the number of moles and the molar mass of mercury. Molar mass of mercury (Hg) = 200.59 g/mol \[0.0272 \text{ mol} \times 200.59 \frac{\text{g}}{\text{mol}} = 5.45 \text{ g}\]

Calculate volume of liquid mercury

Finally, we will find the volume of liquid mercury using the mass of mercury vapor and its density. \[\frac{5.45 \text{g}}{13 \frac{\text{g}}{\text{mL}}} = 0.419 \text{ mL}\] The room will saturate when approximately 0.419 mL of mercury has evaporated and spread throughout the room.