The following data was obtained: atmospheric pressure: 752 torr water vapor pressure: 22 torr height of...

Question

Question

The following data was obtained:

atmospheric pressure: 752 torr

water vapor pressure: 22 torr

height of water column: 158 mm

Calculate the partial pressure of the dry gas. Density of mercury if 13.6 g/cm3 and assume the desnity of the solution is 1.00 g/cm3/

science chemistry

Answer 1

Answer #1

Given that P_atm =752 torr

P_H2O vapor = 22 torr

Height of water column = 158 mm

Density of mercury = 13.6 g/cm³

Density of the solution = 1.00 g/cm³

We know that: P_atm = P_gas + P_H2O vapor

P_gas = 752 torr – 22 torr = 730 torr = 730 mm Hg

If the pressure of the gas (P_gas) is less than the atmospheric pressure then the pressure of dry gas = P_gas – height of mercury column.

Height of water column/height of mercury column = density of mercury/density of water

Height of mercury column = (158 mm x 1 g/cm³)/13.6 g/cm³ = 11.6 nm Hg

Pressure of dry gas = 730 mm Hg – 11.6 nm Hg = 718.4 mm Hg

Answer 2

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