3. (a) A 1 liter closed vessel is contains 400 mls of water and the remaining sp

Question

3. (a) A 1 liter closed vessel is contains 400 mls of water and the remaining space is filled with air. Benzene in an amount of 0.2 gr is added to the vessel. Assuming equilibrium is reached, find the amount (in grams) of benzene in the air and in the water. The solubility of benzene in water is 1.78 g/L. The air(1)-water(2) partition coefficient for benzene is KA12 pAI/PA2 0.22 (b) Suppose that instead of 0.2 gr of benzene, 1.0 gr of benzene was added. What would the amounts of benzene in the water and air be for this case? Would there be any pure benzene? If so, how much?

Explanation / Answer

Volume of water = 0.4 L

Volume of air = 0.6 L

Solubility of benzene in water = 1.78 g/L

Here we have 0.4 L so, max solubility of 1.78 g/L * 0.4 L = 0.712 g

This is Maxx value.

But we have only 0.2 g .

Mass of benzene in air / mass of benzene in water = 0.22

A/W = 0.22

A+W = 0.2

On solving ,

W = 0.180 g

A = 0.02 g

--------------------------

Now we have 1 g of benzene .

Maxx of 0.712 g is possible now.

A/0.712 = 0.22

A = 0.156 g

W = 0.712 g

Remaining = 1-(0.156+0.712) = 0.131 g of pure benzene

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