Question

A mixture of ethyl iodide (c2h5i, bp 72.3°c) and water boils at 63.7°c. what weight of ethyl iodide would be carried over by 1 g of steam during steam distillation?

Answer 1

To determine the weight of ethyl iodide carried over by 1 g of steam during steam distillation, we need to calculate the mole fraction of ethyl iodide in the mixture and multiply it by the weight of the steam.

Step-by-step explanation:

The boiling point of a solution can be determined using Raoult's law, which states that the vapor pressure of a solvent above a solution is equal to the product of the mole fraction of the solvent and the vapor pressure of the pure solvent. In this case, we are looking for the weight of ethyl iodide carried over by 1 g of steam during steam distillation. Since the mixture of ethyl iodide and water boils at 63.7°C, we know that the vapor pressure of ethyl iodide at 63.7°C is equal to the vapor pressure of steam at that temperature.

To determine the weight of ethyl iodide carried over by 1 g of steam, we need to calculate the mole fraction of ethyl iodide in the mixture. We can use the equation:

mole fraction of ethyl iodide = weight of ethyl iodide / weight of ethyl iodide + weight of water

Once we have the mole fraction of ethyl iodide, we can multiply it by the total weight of the steam (1 g) to find the weight of ethyl iodide carried over.

Answer 2

You only need to know first is the vapor pressure of water at that boiling point which is ~ 179 mm hg.

So the (vapor pressure of C2H5I = 760 – 179)

And since you know the molecular weight of H2O which is equal to18 g/mol and the molecular weight of C2H5I = 156 g/mol.

Just apply the equation:

grams H2O M.W. H2O × P H2O

---------------------- = -------------------------- = 28.13 grams C2H5I

grams C2H5I M.W. C2H5I × P C2H5I