Iodobenzene Steam Distills At 98.2

Iodobenzene Steam Distills At 98.2

Iodobenzene (C6H5I, bp 188 ° C) steam is distilled at 98.2 C. How many water molecules are needed for transportation? ۔

Iodobenzene (C6H5I, bp 188 ° C) steam is distilled at 98.2 C. How many water molecules are needed to transfer 1 molecule of iodobenzene? How many grams of iodobenzene per gram of water?

The total vapor pressure of a mixture of two indivisible components is the sum of the partial pressures of each component.

p (t) = p (H2O) + p (Ibenz)

At 98.2 C, the water vapor pressure is 960 mbar. The air pressure (at the boiling point) is p (t) = 1013 mbar. Therefore (1013 960) mBar = 53 mbar iBenzol for partial pressure.

Under ideal gas conditions, we can set partial pressure equivalents or molecules. So we have 960 water molecules that carry 53 ibenzene molecules. 18.1 H2O molecule per iodobenzene molecule.

To convert to grams, multiply by the appropriate values: 18 g / mol and 204 g / mol.

1.6 g H2O per g benzene.

Iodobenzene Steam Distills At 98.2