A student dissolved 45 23 mg of ascorbic acid in enough water to make 100.0 mL o

Question

A student dissolved 45 23 mg of ascorbic acid in enough water to make 100.0 mL of solution. The student then titrated 10.00 mL of this solution with an iodine solution, using starch as the indicator. It required 7.53 mL of the iodine solution to react completely with the sample of the ascorbic acid solution. The student then took a 10.00 mL sample of a fruit juice and titrated it in a similar fashion. The fruit juice required 4.88 mL of the iodine solution for complete reaction. How many grams of ascorbic acid are present in 100 mL of the fruit juice? 69.8 g 29.3 g 6.98 0.0293 g 0.00328 g 0.5764 g 57.64 g 0.328 g

Explanation / Answer

m = 45.23 mg of acid; V = 100 mL

then, V = 10 mL of that solution is titrated with Iodine

V = 7.53 mL of iodine to react

ascorbic acid + I2 --> 2 I- + dehydroascorbic acid

so ratio is 1:1

MW acid = 176.12

mol of acid = Mass/MW = (45.23*10^-3)/176.12 = 2.56813*10^-4 mol of acid..

in 100 mL

we used only 10! so tehre is 10% of it...  2.56813*10^-4*10^-1 =  2.56813*10^-5 mol of acid in 10 mL

so there is  2.56813*10^-5 mol of Iodine

[I2] = mol/V = ( 2.56813*10^-5)/(7.53*10^-3) = 0.003410 M

now...

for the second sample...

we used V = 4.88 mL

mmol = MV = (0.003410)*4.88 = 0.0166408 mmol of Iodine used

since ratio is 1:1

then

0.0166408 mmol of acid (in 10 mL9

for

100 mL juice

0.0166408 --> 10*0.0166408 = 0.166408 mmol of acid

mass = mol*MW = (0.0166408*10^-3)(176.12) = 0.00293 grams

answer is D

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