Question

Given the following solubilities in grams/100ml of water (these are the amounts that stay in solution at these temperatures):

caffeine 1.35g (16 degrees Celcius) 45.5g (65 degrees celcius)

NaCl 35.7g (0 degree celcius) 39.1g (100 degrees celcius)

Suppose that 20g of a mixture that is 75% caffeine and 25% NaCl is crystallized from 100mL of boiling water and cooled to 16 degrees celcius.

a) how much of the caffeine from this mixture stays in solution at 16 degrees celcius?

b) what is the most caffeine from this mixture that can be expected to cystallize?

c) How much NaCl from the original mixture stays in solution?

Answer 1

Given :solubility of caffeine 1.35g at 16⁰ C and 45.5g at 65⁰ C

simillarly

solubility of NaCl 35.7g at 0⁰ C and 39.1g at 100 ⁰ C

Suppose that 20g of a mixture that is 75% caffeine and 25% NaCl is crystallized from 100mL of boiling water and cooled to 16⁰C

Then the quantity of caffeine dissolve in 100 ml water at 100⁰ C = 20 x 75 / 100 = 15 g

the quantity of NaCl dissolve in 100 ml water at 100⁰ C = 20 x 25 / 100 = 5 g

a) as we know the solubility of caffine and NaCl at 100⁰ C are more than total quanty used.

therfore at 100⁰ C a mixture will be cler solution then

mass cool to 16⁰ C at this temp solubility of caffine is 1.35g only

hence quantity of caffine in solution = total quantity of caffine - 1.35

= 15 - 1.35

quantity of caffine in solution at 16⁰C = 13.65g

b) The most of caffeine from this mixture that can be expected to cystallize is 13.65 gm

c) The solubility of NaCl in 100 ml water at 0⁰C is 35.7g however we use 5.0g NaCl for mixture hence total 5.0 gm NaCl remains in solution form.