I dont understand how to do #1. Could someone help explain this and show steps?

Question

I dont understand how to do #1. Could someone help explain this and show steps?

Thanks!

Exp a pest Las you can put draft of calculation on the left side of your notebook. . Suppose you have a substance with a water solubility is 25 g100 mL at 100°C, and Note: 5 g /100 mL at 0°C. (2 points) If you have a total of 15 g of the substance, what volume of water will be necessary to completely dissolve it at 100° C? (1.5 points) What mass of the substance will remain dissolved in the water when it is cooled to 0°C? a. b. c. (1.5 points) What percent of the substance cannot be recovered by recrystallization in water? (2 points) Suppose you were attempting to recrystallize this substance in water, but instead of using the minimal volume of water necessary to dissolve it, you instead added 150 mL of water. Under those circumstances what mass of the substance could you recover upon cooling, and what mass would be unrecoverable? d. 2. (2 points) In recrystallization, the amount of solvent used should be minimized because the more solvent you use, the more product will dissolve in the cold solvent and will be lost. Ideally, it should be a saturated solution in hot solvent. Acetanilide has a solubility of 5.3g/100g of 100°C of water, based on your theoretical yield in the experiment, what is the minimum amount of water you need in this experiment? 3. (3 points) In a two-step synthesis, anthranilic acid(CH NO2) is acetylated with acetic anhydride to give N-acetylanthranilic acid(CoHNO2), and then form 2- methylbenzisoxazinone(CoHoNOs),. supposing you have 2.00g of anthranilic acid, the yield of first step is 80% and the second step is 60%. What is the weight of

Explanation / Answer

Ans. #a. Required volume of water = Mass of solute / Solubility at 100.00C

                                                            = 15 g / (25 g / 100 mL)

                                                            = 60 mL

#b. Given, solubility at 0.00C = 5 g / 100 mL

When the solution is gradually cooled, all the solute above the specified solubility at colder temperature is precipitated. In other terms, when cooled, the solution must have a concentration of 5 g/ 100 mL.

So,

            Amount of solute remaining in solvated form when the above solution (#a) is cooled from 1000C to 00C = Vol. of soln. x Solubility at 00C

                                                = 60 mL x (5 g / 100 mL)

                                                = 3 g

#c. % not recovered = (Mass of unrecovered solute / Total sample mass) x 100

                                    = (3 g / 15 g) x 100

                                    = 20 %

#d. Concentration of new solution = 15 g / 150 mL                  - at 1000C

            Amount of solute remaining in solvated form when the above solution (#d) is cooled from 1000C to 00C = Vol. of soln. x Solubility at 00C

                                                = 150 mL x (5 g / 100 mL)

                                                = 7.5 g

So, the unrecoverable mass of solute = 7.5 g

# Mass of recovered solute = Total mass of solute taken – Mass of unrecovered solute

                                                = 15 g – 7.5 g

                                                = 7.5 g

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