Question

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?

a) 9.63 × 10^⁻4 torr
b) 2.19 × 10^⁻3 torr
c) 4.38 × 10^⁻3 torr
d) 8.75 × 10^⁻3 torr

Answer 1

The partial pressure of gaseous mercury can be calculated using the ideal gas law equation and the given concentration. The partial pressure is approximately 9.63 x 10^-4 torr.

Step-by-step explanation:

To find the partial pressure of gaseous mercury, we first need to convert the given concentration from mg/L to torr. The molar mass of mercury is 200.59 g/mol. Using the ideal gas law equation, we can calculate the partial pressure P of the gaseous mercury:

P = (n/V) * RT

where n is the number of moles of gaseous mercury, V is the volume of air, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the equation, we get:

P = (c/M) * RT

where c is the concentration in mg/L, M is the molar mass of mercury, and R is the ideal gas constant. Plugging in the values for c, M, R, and T, we can calculate the partial pressure of gaseous mercury.

Using the given concentration of 2 x 10^-6 mg/L and the molar mass of mercury, the partial pressure of gaseous mercury is approximately 9.63 x 10^-4 torr.