We’re being asked to calculate the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 °C.
First. we’re going to calculate the total concentration of air using the ideal gas equation:
P = pressure, atm
V = volume, L
n = moles, mol
R = gas constant = 0.08206 (L·atm)/(mol·K)
T = temperature, K
Ptotal = 733 torr → convert to atm
Ptotal = 0.9645 atm
R = 0.08206 (L∙atm/(mol∙K)
T = 26°C + 273.15 = 299.15 K
Recall: molarity (M) = mol/L or n/V
Rearranging the ideal gas equation:
A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as 2 × 10−6 mg/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 °C?
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