Question

Question: The solubility of compound A in water at 25C is 3.00g/100mL. In benzene at the same temperature, the solubility of compound A is 4.00g/50mL. If 3g of compound were dissolved in 100mL of water, how much compound would be extracted by three 10mL portions of benzene?

Answer 1

Answer:

To determine how much compound would be extracted

In water,

Solubility of product = 3g/100 ml

In benzene,

Solubility of compound = 4 g/50 ml

= 8 g/100 ml

We know,

K = [A]org / [A]water

where as K is partial coefficient between water and benzene

substitute values

K = (8 g/100 ml)/ (3 g/ 100 ml)

K = 2.6667

For 10 ml of C6H6(Benzene)

Fraction of solute in water = q

q = (Veq / (KdVorg + V aq)^n ---------->(1)

where as

n is number of extractions= 3

substitute the given values in the expression (1)

q = (100/ (2.6667*10) + 100)^3

q = (100/126.67)^3

q = 0.4921

Compound remain in the aqueous phase = 0.4921*3 g

= 1.48 g

Compound extracted with C6H6 i.e., benzene = 3- 1.48

= 1.52