Question

the solubility of a compound is 59 g / 100 ml in boiling methanol and 30 g / 100 ml in cold methanol. how much mass would you recover if you take 7.2g of compound and dissolve it in 100 ml of boiling methanol and then cool it down to 2°c and filter it.

Explanation:

Step1: Determine solubility at cold temperature

The solubility of the compound in cold methanol is 30 g/100 mL. When 100 mL of the solution is cooled, the amount of compound that remains dissolved is based on this cold - temperature solubility.

Step2: Calculate the amount of undissolved compound

We started with 7.2 g of the compound in 100 mL of boiling methanol. The amount of compound that remains dissolved in 100 mL of cold methanol is 30 g (from the solubility data). The amount of compound that will precipitate out (be recovered) is the initial amount minus the amount that remains dissolved.
Let $m$ be the mass of the recovered compound.
$m=7.2 - 30\times\frac{7.2}{59}$
First, calculate $\frac{7.2}{59}\approx0.122$. Then $30\times0.122 = 3.66$ g.
$m = 7.2-3.66=3.54$ g

Answer:

3.54 g