The solution of Malonitrile in ether is 30 g per 100 mL and in water @ room temp

Question

The solution of Malonitrile in ether is 30 g per 100 mL and in water @ room temperature is 19.95 g per 100 mL. Suppose a reaction mixture when diluted with water afforded 500mL of an aqueous solution of 25 g of the reaction product malonitrile, CH_2(CH)_2 which is to be isolated by extraction with ether. a) What is K_D -- Distribution coefficient? b) What Weight of Malonitrile would be recovered by extraction with one-500mL of ether? - SHOW YOUR CALCULATION CLEARLY c) What weight of Malonitrile would be recovered by 4-12SmL of ether? SHOW YOUR WORK CLEARLY

Explanation / Answer

(a)

Kd = [conc in organic solvent]/[conc in aqueous solution]

So, putting values we get:

Kd = [30/100]/[19.95/100] = 1.503

When extraction with 500mL ether is performed, the amount of solute in ether would be 1.503 times more than that in 500ml water.

(b)

Assuming mass of solute that goes into ether solvent as 'x' grams, we have:

x + x/1.5 = 25

Solving we get:
x = 15 g

(c)

Assume mass of solute in ether as 'x' g and in water as 'y' g, after first extraction of 125 mL ether, we have:

(x/125)/(y/500) = 1.503

So,

x/y = 0.375

And

x + y = 25

Thus, solving above we get:

x = 6.8 g

Amount of solute remaining = 25-6.8 = 18.2 g

Similarly for next extraction, we have the following equations:

x/y = 0.375

And

x + y = 18

So, x = 4.96

So, mass of solute remaining = 18.2 - 4.96 = 13.24 g

Similarly, for third extraction we have, x = 3.61 g

So, mass of solute remaining = 13.24 - 3.61 = 9.63 g

Finallly, for fourth extraction we have, x = 2.62 g

So, solute remaining = 9.63-2.62 = 7.01 g

So, total solute extracted after four extractions of 125mL each = 25-7.01 = 17.99 g

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