2. A XXXXXXXXXXg sample of pure ascorbic acid is weighed out and dissolved in 50.0 mL of distilled...

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2. A XXXXXXXXXXg sample of pure ascorbic acid is weighed out and dissolved in 50.0 mL of distilled water in an Erlenmeyer flask. One week later, when about one third of the water in the flask has evaporated, this solution is titrated against XXXXXXXXXXM K103 solution and requires 18.75 mL of it to reach the end point.
(a) What mass of ascorbic acid has been oxidized by the air over the week-long waiting period?

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Robert · Accepted Answer

Solution:TTS171112_51130_28
0.1967 g ascorbic acid = 0.1967/176.12 moles ascorbic acid = 0.001117 moles 
We know: 
Moles of solute = volume of solution used (in Liter) x molarity of the solution 
So, 18.75 mL(= 0.01875L) of 0.01735 M KIO3 solution contains                                               
( 0.01875L) x (0.01735 moles/L) =  0.0003253 = moles of KIO3.

2. A XXXXXXXXXXg sample of pure ascorbic acid is weighed out and dissolved in 50.0 mL of distilled water in an Erlenmeyer flask. One week later, when about one third of the water in the flask has...

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