The minimum volume of water needed is 0.28 mL. The solution will have
1.7 mg of phthalic acid remaining in solution.
At 100 °C
Volume of water required = 50 mg HPh × (100 mL water/18 000 mg HPh)
= 0.28 mL water
At 25 °C
Mass of HPh in solution = 0.28 mL water × (620 mg HPh/100 mg water)
= 1.7 mg HPh
If Phthalic acid has a solubility of approximately 18g per 100 ml of water at 100°c and 0.62g per 100 ml of water at 25°c.
∴
The minimum volume of water needed to recrystallize 50 mg of phthalic acid is 0.28 mL.
Also, the amount of phthalic acid remaining after the solution recrystallizes is 48.264 g.
From the parameters given:
At 100° C, if 100 mL could dissolve 18 g Phthalic acid;
∴
1 mL of water will be able to dissolve = (1 mL × 18 g)/100 mL
= 0.18 g of phthalic acid.
Thus, the minimum volume to dissolve 50 mg of phthalic acid would be:
x = (0.05 g × 100 mL)/18 g
x = 0.28 mL
It implies that 0.28 mL will be the minimum volume to dissolve 50 mg of phthalic acid when it cools before it starts to recrystallize.
Similarly, at 25° C, 0.62 g was able to be dissolved in 100 mL of water.
∴
1 mL of water will be able to dissolve = (1 mL × 0.62 g)/100 mL
= 0.0062 g of phthalic acid.
= 6.2 mg of phthalic acid
Thus,
0.28 ml of water from the 50 mg of phthalic acid will be able to dissolve:
= (6.2 mg × 0.28 mL)/1 mL
= 1.736 mg
This implies that the moment the phthalic acid crystalize from the solution;
1.736 mg was able to be dissolved and the remaining amount of phthalic acid undissolved after recrystallization is:
= 50 mg - 1.736 mg
= 48.264 mg
Therefore, we can conclude that the minimum volume of water needed to recrystallize 50 mg of phthalic acid is 0.28 mL and the amount of phthalic acid remaining after the solution recrystallizes is 48.264 g.
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