A 500.00 mg vitamin C (MW176.12g/mol) tablet was ground, acidified, and dissolved in H2C to make a 250.0 mL solution. A 50.00 mL aliquot containing vitamin C, KI and starch was analyzed and titrated with 12.31 mL of 0.01042 M KIO3. What is the % (w/w) vitamin C in the tablet? KIO3 + 5 KI + 6 H* = 3 12 + 6 K* + 3 H20 C6H8O6 + 12 = C6H606 + 21 + 2H* 3.39 % 33.89 % 67.77 %

A 500.00 mg vitamin C (MW176.12g/mol) tablet was ground, acidified, and dissolved in H2C to make a 250.0 mL solution. A 50.00 mL aliquot containing vitamin C, KI and starch was analyzed and titrated with 12.31 mL of 0.01042 M KIO3. What is the % (w/w) vitamin C in the tablet? KIO3 + 5 KI + 6 H* = 3 12 + 6 K* + 3 H20 C6H8O6 + 12 = C6H606 + 21 + 2H* 3.39 % 33.89 % 67.77 %

Introductory Chemistry For Today

8th Edition

ISBN:9781285644561

Author:Seager

Publisher:Seager

Chapter7: Solutions And Colloids

Section: Chapter Questions

Problem 7.34E

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A 500.00 mg vitamin C (MW176.12g/mol) tablet was ground, acidified, and dissolved in H2O to make a 250.0 mL solution. A 50.00 mL aliquot containing vitamin C, KI and starch was analyzed and titrated with 12.31 mL of 0.01042 M KIO3. What is the % (w/w) vitamin C in the tablet? KIO3 + 5 KI + 6 H+ = 3 I2 + 6 K+ + 3 H2O C6H8O6 + I2 = C6H6O6 + 2I- + 2H+ 67.77 % 3.39 % 22.59 % 33.89 %
A 500.00 mg vitamin C (MW176.12g/mol) tablet was ground, acidified, and dissolved in H2O to make a 250.0 mL solution. A 50.00 mL aliquot containing vitamin C, KI and starch was analyzed and titrated with 12.31 mL of 0.01042 M KIO3. What is the % (w/w) vitamin C in the tablet? KIO3 + 5 KI + 6 H+ = 3 I2 + 6 K+ + 3 H2O C6H8O6 + I2 = C6H6O6 + 2I- + 2H+ A 67.77 % B 33.89 % C 22.59 % D 3.39 %
A 15.0% by weight solution was prepared using 90.0g of KCl and the resulting density of the solution is 1.101g/mL. (MW KCl 74g/n). volume of solution(mL)? milliosmole of solute? %w/v? weight of solvent(g? N?
A 500-mg tablet of commercial vitamin C tablet was dissolved in 50 mL water, treated with excess KI solution and 3 drops of starch solution. The solution required 4.8 mL of 0.05 M KIO3 to reach the endpoint. Calculate the % by weight of vitamin C in the tablet. (Molar mass of vitamin C = 172.16 g mol-1) IO3‐ + 5 I‐ + 6 H+ → 3 I2 + 3 H2O (1) I2 + C6H8O6 → C6H8O6* + 2 H+ + 2 I‐ (2) I2 + starch → (I2 - starch) (deep blue color) (3) *oxidized product of vitamin C
7 of 26 > Calculate the percent mass per volume, % (m/v), of a dextrose solution containing 6.50 g of dextrose in 2.00 × 10² mL of solution. Note that mass is not technically the same as weight, but the abbreviation % (w/v) is often used interchangeably with % (m/v). mass/volume %: % (m/v)
A mixture contains Na2CO3, NaOH and inert matter. A sample weighing 1.500 g requires 28.85 ml of 0.500N HCI to reach a phenolphthalein endpoint, and an additional 23.85 to reach a methyl orange endpoint.What are the percentages of Na2C03 and NaOH?
3. Vida and Min dissolved 0.0521 g of C6H8O6 in a 100 mL volumetric flask with deionized water. A 10.0 mL aliquot of this standard CHO solution required 23.7 mL of DCIP to reach the end point. A 5.00 mL aliquot of orange juice required 14.4 mL of the same DCIP solution to reach the end point. a. What is the molarity of the standard ascorbic acid solution? b. What is the molarity of the DCIP solution? c. How many mg of C6H8O6 are present in the juice sample? d. How many mL of orange juice are needed to satisfy the adult DRI requirement for Vitamin C?
Calculate the molar concentration of a thiosulfate solution from the following information:A 40.-mL aliquot of a 0.00653 M KIO3 solution is added to a flask containing 2 g of KI and 10 mL of 0.5 M H2SO4. The resulting solution is titrated to a starch endpoint with 38 mL of the thiosulfate solution.
Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample. The stated should be 104.5, how??
Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample.
On an emergency cart, you have sodium bicarbonate solution (NaHCO3), 44.6 mEq/50 ml. A physician orders an aerosol of 5 cc and 3.25% strength. How many milliliters of the bicarbonate solution do you need? 1 mEq=1/1000 GEW; GEW= gram formula wt/valence Atomic weights: Na, 23; H, 1; C, 12; 0, 16
The weight percent %W/W of a saturated solution of potassium chloride in 200CO was: (NOTE: the solution contain 298g/L and D=1.17g/cm3) * 40.66%W/W W/W% 75.20 W/W% 25.47 10.54%W/W