Question

A 3 0.0 gram mixture containing 88.0% (by weight) of A and 12.0 % of B is recrystallized in water. The solubility of A and B in 100 mL of water is as follows:

   20 ̊ C       100 ̊C

A    1.5 g        6.0 g

B    0.75 g     4.0 g

a. Assuming that A and B do not affect each other’s solubility, c alculate the amount of water (in mL) that would be needed to obtain pure A . Show all work. Explain answers.

b. How much of pure A will be recovered? Show all work. Explain answers.

Answer 1

As mentioned in question

a). Assuming that A and B do not affect each other’s solubility, c alculate the amount of water (in mL) that would be needed to obtain pure A . Show all work. Explain answers.

answer

A 3 0.0 gram mixture containing 88.0% (by weight) of A ,

so it 30 g * 88

A =26.4 g

12.0 % of B , So 30*12

B = 3.6 g

The solubility of A and B in 100 mL of water is as follows:

   20 ̊ C       100 ̊C

A    1.5 g        6.0 g

B    0.75 g     4.0 g

Recrystalisation means it sholud be complete soluble at hot condition, agin reform solids at cooling temperature.

so hot condition solubility for A is 6 g in 100 ml

we have to purify 26.4 g (as abovecalculated 88% in 30 g) of A, water required?

for 6 g required 100 ml water

26.4 g required water ?

=26.4 *100/6

=440 ml

so 440 ml required for A purification.

b).  How much of pure A will be recovered? Show all work. Explain answers.

Answer:

same above mentioed solubility

The solubility of A and B in 100 mL of water is as follows:

   20 ̊ C       100 ̊C

A    1.5 g        6.0 g

in recrystalization we have to cool the reaction mass, so at cooling 1.5 g in 100 ml.

but total we have 26.4 g and using 440 ml of water (above a answer),

so in 440 ml loss

=440*1.5/100

= 6.6 g would be loss due to its solubility (it present in filter mls water).

so the amount of pure compound will get after recrystalization

= 26.4 g -6.6 g

=19.8 g.

pure compound of A will recover 19.8 g