Question

In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution three times with 5.0-ml portions of methylene chloride. calculate the total amount of caffeine that can be extracted into the three portions of methylene chloride (see technique 12, section 12.2). caffeine has a distribution coefficient of 4.6 between methylene chloride and water.

Answer 1

solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

No of portions n=3

Distribution co-efficient= 4.6

Total amount of caffeine that can be unextracted is given by

$w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\

w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\

=0.170[\frac{46}{46+5}]^3\\

=0.170[\frac{46}{51}]^3\\

=0.170[\frac{97336}{132651}]\\

=0.170\times0.734=0.125gms$

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

                                                       =0.045 gms