THE AMOUNT OF VITAMIN C IN FOOD Eq1: ( 6H+) + (5I-) >> 3 I2 + 3H2O Eq2: C6H8O6+I
ID: 917897 • Letter: T
Question
THE AMOUNT OF VITAMIN C IN FOODEq1: ( 6H+) + (5I-) >> 3 I2 + 3H2O Eq2: C6H8O6+I2>>C6H6O6+(2H+)+(2I-) Eq3: 3C6H8O6+KIO3 >> 3C6H6O6 +KI +3H2O Eq4: I2 + I- >> I3-
Use the volume of titrant used V= 0.0327 L, the concentration, the mole ratio between KIO3 and ascorbic acid from (eq 3), and the mass of ascorbic acid to CALCULATE the molecular weight of ascorbic acid (expt' l) . How did you that? Please explain the step by step.
Mass of ascorbic acid : 202g Final buret level 32.7 ml Initial buret level : 0.00ml Volume of titrant (KIO3) used: 32.7 ml Concentration of KIO3 (IO3-) solution: 0.0122M. Mass of KIO3 : 0.651g Molecular weight of KIO3: 214.0028g/mol THE AMOUNT OF VITAMIN C IN FOOD
Eq1: ( 6H+) + (5I-) >> 3 I2 + 3H2O Eq2: C6H8O6+I2>>C6H6O6+(2H+)+(2I-) Eq3: 3C6H8O6+KIO3 >> 3C6H6O6 +KI +3H2O Eq4: I2 + I- >> I3-
Use the volume of titrant used V= 0.0327 L, the concentration, the mole ratio between KIO3 and ascorbic acid from (eq 3), and the mass of ascorbic acid to CALCULATE the molecular weight of ascorbic acid (expt' l) . How did you that? Please explain the step by step.
Mass of ascorbic acid : 202g Final buret level 32.7 ml Initial buret level : 0.00ml Volume of titrant (KIO3) used: 32.7 ml Concentration of KIO3 (IO3-) solution: 0.0122M. Mass of KIO3 : 0.651g Molecular weight of KIO3: 214.0028g/mol THE AMOUNT OF VITAMIN C IN FOOD
Eq1: ( 6H+) + (5I-) >> 3 I2 + 3H2O Eq2: C6H8O6+I2>>C6H6O6+(2H+)+(2I-) Eq3: 3C6H8O6+KIO3 >> 3C6H6O6 +KI +3H2O Eq4: I2 + I- >> I3-
How did you that? Please explain the step by step.
Mass of ascorbic acid : 202g Final buret level 32.7 ml Initial buret level : 0.00ml Volume of titrant (KIO3) used: 32.7 ml Concentration of KIO3 (IO3-) solution: 0.0122M. Mass of KIO3 : 0.651g Molecular weight of KIO3: 214.0028g/mol
Mass of ascorbic acid : 202g Final buret level 32.7 ml Initial buret level : 0.00ml Volume of titrant (KIO3) used: 32.7 ml Concentration of KIO3 (IO3-) solution: 0.0122M. Mass of KIO3 : 0.651g Molecular weight of KIO3: 214.0028g/mol
Explanation / Answer
The reactions will be
IO3- + 5I- + 6H+ --> 3I2 + 3H2O
Total moles of IO3- used = volume X molarity = 32.7 X 0.0122 = 0.398 millimoles
Iodine produced = 3 X 0.398 millimoles = 1.194 millimoles
Vitamin C + I2 --> 2I-
Moles of vitamins c used = 1.194 millimoles
Mass = 202 grams
so molecular weight = mass / moles = 202 / 1.194 millimoles = 169.17 g / mole
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