A student purified a 3.88 g sample of phthalic acid by recrystallization from water. The published solubility of phthalic acid in 100 mL of water is 0.54 g at 14ºC and 18 g at 99ºC. What is the smallest volume (in milliliters) of boiling water the student could use to dissolve the 3.88 g sample of phthalic acid? (Report your answer to 3 significant figures.)
100 mL x 3.88g/18g = ?? mL at 99 C. We will assume that is at 100.
718
13.2ml
To find the smallest volume of boiling water needed to dissolve the 3.88 g sample of phthalic acid, we can use the solubility data given. The solubility of phthalic acid in water increases with temperature, so we need to find the solubility at the highest temperature provided (99ºC).
We can set up a proportion to find the volume of water needed.
Let's use x to represent the volume of boiling water (in mL).
The proportion can be set up as:
0.54 g/100 mL = 18 g/x mL
Cross-multiplying gives us:
0.54*x = 18*100
0.54x = 1800
Now, we can solve for x:
x = 1800/0.54
x ≈ 3333.33 mL
Rounding to three significant figures, the smallest volume of boiling water needed to dissolve the 3.88 g sample of phthalic acid is approximately 3330 mL.