A $12.0 g$ sample contains $90%$ of compound A and $10%$ of an impurity that stays dissolved throughout. The solubility of A is $24.0 g /100 mL$ in a solvent at its boiling point and $3.0 g /100 mL$ at $5^{\circ} C$ . The sample is dissolved in the minimum volume of boiling solvent and cooled to $5^{\circ} C$ . The mass of A crystallised is

Answer: A

Mass of A

= 12.0 \times 0.90 = 10.80

g. Minimum solvent

= 10.80/24.0 \times 100 = 45.0

mL. A retained at

5^{\circ} C

= 45.0 \times 3.0/100 = 1.35

g. Crystals of A

= 10.80 - 1.35 = 9.45