The distribution coefficient, KD = (Cether)/(Cwater) = 3.5 for aspirin at 25 C. What volume of ether would be necessary to extract 0.95 g of aspirin from a solution of 1.0 g of aspirin in 100 mL of water in a single extraction? Show all calculations.
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The distribution coefficient, KD = (Cether)/(Cwater) = 3.5 for aspirin at 25 C. What volume of ether would be necessary to extract 0.95 g of aspirin from a solution of 1.0 g of aspirin in 100 mL of water in a single extraction? Show all calculations.
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- The distribution coefficient, KD (ether/water), between ether and water for aspirin at room temperature is 3.5. What weight of aspirin would be extracted by a single extraction with 60 ml of water from a solution of 10 grams of aspirin in 100 ml of ether? Calculate the weight of aspirin which would be removed by three extractions with 20 ml portions of water.The distribution coefficient, KD (C ether/C water), between ether and water for aspirin at room temperature is 3.5. What weight of aspirin would be extracted by a single extraction with 150 mL of water from a solution of 10 gms of aspirin in 100 mL of ether? Calculate the weight of aspirin which would be removed by three extractions with 50 mL portions of water.If 500 mL of a 15% (v/v) solution of methylsalicylate in alcohol is diluted to 1500 mL, what will be the percentage strength (v/v)? a 50% b 60% c 6% d 5% e 10%
- 1. Why should the stopper always be removed from a separatory funnel whenever a liquid is being drained through the stopcock?2. The distribution coefficient, KD (ether/water), between ether and water for aspirin at room temperature is 3.5. What weight of aspirin would be extracted by a single extraction with 60 ml of water from a solution of 10 grams of aspirin in 100 ml of ether? Calculate the weight of aspirin which would be removed by three extractions with 20 ml portions of water.How efficient is the extraction of caffeine from tea leaves containing 1.0 g caffeine in 80 ml aqueous extract with two 30-ml portions of dichloromethane over that of a single-step extraction (K at 25°C = 8.36)?A 25 g of butyric acid was extracted with 400 mL ether from 400 mL water (Kd=3.0 at 25 °C) Using single-step extraction, what is the amount of butyric acid in water layer? What is the total amount of butyric acid extracted using two successive extractions? Calculate the % efficiency of two-step extraction over single
- 1. What is the MW of the analyte?2. What is the meq of the analyte?3. What will be the weight (in grams) of the analyte if the standardization is to be quartered?4. What will be Normality of the solution if 7.5ml of the titrant was consumed in the standardization?5. What is the average Normality if two or more trials were conducted with the values of 0.102N and 0.105N?1. What is the MW of the analyte?2. What is the meq of the analyte?3. What is the weight (in grams) of the analyte?4. What will be the percentage purity of the sample if 43mL of the titrant was consumed?5. Did the sample conform to the USP requirement? Yes or No?What is the minimum distribution constant that permits removal of 99% of a solute from 50.0 mL of water with two 25.0-mL extractions with toluene? five 10.0-mL extractions with toluene?
- Q1. Dissolved 0.273 grams of pure sodium oxalate (Na,C,O.) in distilled water and added sulfuric acid and titration the solution at 70 ° C by using 42.68 ml of KMNO, solution and has exceeded end point limits by using 1.46 ml of standard oxalic acid (H; C;O.) with 0.1024 N. Calculate the normlity of KMN0.. Note that the molecular weight of sodium oxalate (Na,C,O.) = 134 and its equivalent weight = 67What is the ratio strength (w/v) of an isotonic solution for pilocarpine hydrochloride FD1% 0.138? A. 1 in 26.5 B. 1 in 37.7 C. 1 in 46.1 D. 1 in 217 Can you explain how to tackle such questions that only provides SCE values or FD1 values like the above? Thank youA scientist wishes to measure the concentration of methyl benzoate in a plant stream by gas chromatography. He prepares a sample of butyl benzoate to use as an internal standard. The results of a preliminary run, which used a solution known to contain 1.37 mg/mL of methyl benzoate (peak A) and 1.51 mg/mL of butyl benzoate (peak B), are shown. The area of peak A is determined to be 3.40×102 and the area of peak B is determined to be 398 measured in arbitrary units by the computer. To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.49 mg/mL is mixed with 1.00 mL of the plant stream material. Analysis of the mixture gave a peak area of 437 for peak A and 415 for peak B. What is the concentration of methyl benzoate in the plant stream?