Question

After the large eruption of Mount St. Helens in 1980, gas samples from the volcano were taken by sampling the downwind gas plume. The unfiltered gas samples were passed over a gold-coated wire coil to absorb mercury (Hg) present in the gas. The mercury was recovered from the coil by heating it and then analyzed. In one particular set of experiments scientists found a mercury vapor level of 1800 ng of Hg per cubic meter in the plume at a gas temperature of 12 ∘C.

Part A: Calculate the partial pressure of Hg vapor in the plume.Express your answer using two significant figures.

Part B: Calculate the number of Hg atoms per cubic meter in the gas.Express your answer using two significant figures.

Part C: Calculate the total mass of Hg emitted per day by the volcano if the daily plume volume was 1700 km3 .Express your answer using two significant figures.

thank you in advance

Answer 1

Given vapor level of mercury(Hg) in sample = 1800 ng of Hg / m³

molar mass of Hg = 200 g Hg/mol

Part A;

dividing vapor level (m/V) with molar mass(M),

((1800 x 10^-9 g Hg) / m³) / (200 g Hg/mol) = 9 × 10^-9 mol/m³ Hg
As,
PV = nRT
P = (n/V) RT = (9x10^-9 mol/m³)x(8.314 m³·Pa/K·mol)x(12 + 273 K) = 2.13x10^-5 Pa
Hence partial pressure of mercury(Hg) = 2.13x10^-5 Pa

Part B;
Now for volume of 1 m³

the mass of Hg in 1 m³ of plume = 1800x10^-9 g

hence number of moles will be,

(1800 x 10^-9 g) / (200 g Hg/mol) = 9x10^-9 mol

number of atoms = (9x10^-9 mol)x (6.022 × 10²³ atoms/mol) = 5.42 x 10¹⁵ atom Hg

Part C;

mass of mercury in 1 m³ of plume = 1800x10^-9 g

now daily quantity of plume = 1700 km³ = 1700x10⁹ m³

total mass emitted per day = 1800x10^-9 g/m³ x 1700x10⁹ m³ = 3.06x10⁶ g = 3.06x10⁶/1000 kg

=3060 kg

hence 3060 kg of Hg is emitted by the volcano each day.