A student breaks a thermometer and spills most of the mercury (Hg) onto the floor of a laboratory that measures 15.2 m long, 6.6 m wide, and 2.4 m high.

Question

A student breaks a thermometer and spills most of the mercury (Hg) onto the floor of a laboratory that measures 15.2 m long, 6.6 m wide, and 2.4 m high.

(a) Calculate the mass of mercury vapor (in grams) in the room at 20°C. The vapor pressure of mercury at 20°C is 1.7 × 106 atm.
(b) Does the concentration of mercury vapor exceed the air quality regulation of 0.050 mg Hg/m3 of air?
(c) One way to treat small quantities of spilled mercury is to spray sulfur powder over the metal. Suggest a physical reason for this action.

Answer 1

Convert the measurements to volume; i.e., v = length x width x height. You will have volume in cubic meters.

Use PV = nRT and solve for n = mol Hg.
Convert mols to grams. g = mols x atomic mass.
That gets a. You can do the comparison for b.
c. Chemical reason might be formation of HgS. Physical might be to serve as a cap or cover for the spill. Over time I don't think that will be effective as the vapor pressure, at equilibrium, will be the same whether covered or uncovered; however, it will take longer to reach equilibrium.

Answer 2

To solve this problem, we need to use the ideal gas law:

PV = nRT

Where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant
T is the temperature

Let's start by calculating the number of moles of mercury vapor using the given information:

(a) Calculate the mass of mercury vapor (in grams) in the room at 20°C:

1. Convert the temperature from °C to Kelvin:
T = 20 + 273.15 = 293.15 K

2. Convert the pressure from atm to Pa:
P = 1.7 × 10^(-6) atm * 101325 Pa/atm = 0.17279 Pa

3. Calculate the volume of the room:
V = 15.2 m * 6.6 m * 2.4 m = 238.464 m^3

4. Rearrange the ideal gas law to solve for the number of moles (n):
n = PV / RT

n = (0.17279 Pa) * (238.464 m^3) / [(8.314 J/mol*K) * (293.15 K)]

n = 0.0176 mol

5. Calculate the molar mass of mercury (Hg):
The molar mass of Hg is approximately 200.59 g/mol.

6. Calculate the mass of mercury vapor:
mass = n * molar mass

= 0.0176 mol * 200.59 g/mol

≈ 3.512 g

Therefore, the mass of mercury vapor in the room at 20°C is approximately 3.512 grams.

(b) Does the concentration of mercury vapor exceed the air quality regulation of 0.050 mg Hg/m3 of air?

To answer this question, we need to convert the concentration of mercury vapor from grams to mg and compare it with the air quality regulation.

The concentration of mercury vapor in the room can be calculated as follows:

concentration = (mass / volume) * 1000

= (3.512 g / 238.464 m^3) * 1000

≈ 14.7 mg/m^3

Since the concentration of mercury vapor in the room is 14.7 mg/m^3, which is higher than the air quality regulation of 0.050 mg Hg/m^3, it exceeds the air quality regulation.

(c) One way to treat small quantities of spilled mercury is to spray sulfur powder over the metal. Suggest a physical reason for this action.

Sulfur powder is often used to treat small quantities of spilled mercury because it forms an insoluble compound called mercury sulfide (HgS) when they react. Mercury sulfide is not volatile and therefore reduces the release of mercury vapor into the air. This action helps to minimize the potential health hazards associated with mercury vapor inhalation.

Answer 3

To answer these questions, we need to first calculate the number of moles of mercury vapor in the room and then convert it to mass. Let's go step-by-step:

(a) To calculate the mass of mercury vapor in the room, we need to use the Ideal Gas Law:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)

First, let's convert the pressure from atm to Pascals (Pa):

1 atm = 101325 Pa

So, the pressure of mercury vapor is:
1.7 × 10^-6 atm = 1.7 × 10^-6 × 101325 Pa = 0.17145 Pa

Next, let's convert the volume of the room from cubic meters (m^3) to liters:

volume = length × width × height = 15.2 m × 6.6 m × 2.4 m = 239.808 m^3

Now, let's convert the temperature from degrees Celsius to Kelvin:

T(K) = T(°C) + 273.15 = 20 + 273.15 = 293.15 K

Now we have all the values to calculate the number of moles:

n = (PV) / (RT) = (0.17145 Pa × 239.808 L) / (0.0821 L.atm/mol.K × 293.15 K) = 2.803 × 10^-3 moles

To find the mass, we need to know the molar mass of mercury (Hg):

Molar mass of Hg = 200.59 g/mol

mass = number of moles × molar mass = 2.803 × 10^-3 moles × 200.59 g/mol = 0.561 g

Therefore, the mass of mercury vapor in the room at 20°C is approximately 0.561 grams.

(b) To determine if the concentration of mercury vapor exceeds the air quality regulation of 0.050 mg Hg/m^3, we need to calculate the concentration.

Concentration = (mass of mercury vapor) / (volume of the room)
= 0.561 g / 239.808 m^3

Let's convert grams to milligrams to match the units in the air quality regulation:

Concentration = (0.561 g × 1000 mg/g) / 239.808 m^3
= 2.338 mg/m^3

The concentration of mercury vapor in the room is 2.338 mg/m^3, which exceeds the air quality regulation limit of 0.050 mg Hg/m^3.

(c) Spraying sulfur powder over the metal is a suggested treatment for small quantities of spilled mercury because sulfur reacts with mercury to form a black mercury sulfide (HgS) precipitate. This reaction renders the mercury less toxic and facilitates its cleanup. The black mercury sulfide is more stable and less likely to release toxic mercury vapor into the air.