I need questions 1 through 3 answered, PLEASE. Here\'s what I need to know, back

Question

I need questions 1 through 3 answered, PLEASE. Here's what I need to know, background info is below AFTER my questions :

I NEED TO KNOW, ASAP:

1 ) The AVERAGE MOLES of ascorbic acid in the CRUSHED tablet (weight of crushed tablet is 1.4844 g)

2 ) Calculate the MASS of ascorbic acid in the crushed tablet. *Company claim is 500mg ascrorbic acid / tablet

3 ) Calculate the MASS of ascorbic acid in the WHOLE tablet (whole tablet, before crushed was 1.5207 g)

WHAT I ALREADY KNOW:

A:

1gm of KIO3 contain =9.5 mM

1.0838 gm will contain = 9.5*1.0838=10.296 mM

Concentration : no. of moles of KIO3/ Volume of solvent= 10.2961 mM/0.5 L =2.05922 * 10^-2 moles/Litres (ans)

B:

1st trail

no.of moles of KIO3 =2.05922*10^-2 * 0.01807=3.721 *10^-4 moles

2nd trail

no.of moles of KIO3 =2.05922*10^-2 * 0.01808=3.723 *10^-4moles

Average No. of moles of Iodate ion added =3.722 *10^-4 moles

C:

Average No. of moles of Iodate ion added =3.722 *10^-4 moles= average no. of moles of ascorbic acid in the pipetted sample (As at equivalence point no. of moles ofIodate ion and ascorbic acid will be same)

Explanation / Answer

Part A

already answered

Part B

Average No. of moles of Iodate ion added =3.722 *10^-4 moles= average no. of moles of ascorbic acid

Mole= weight of the ascorbic acid/ Molecular weight of Ascorbic acid

=>weight of the ascorbic acid= Mole*Molecular weight of Ascorbic acid

=>weight of the ascorbic acid= 3.722*10^-4*176= 655.072*10^-4=6.55*10^-2 gm

So the Mass of ascorbic acid present in the crushed tablet is 65.5 mgm (Ans)

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