Here\'s what I need to know, background info is below : I need to know - 1 ) The

Question

Here's what I need to know, background info is below :

I need to know -

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:

Part 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)

Part 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

Part 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

hope i help!

Part A)
C = 12.011
H = 1.00794
O = 15.9994
Mm of C6H8O6 = 6(12.011) + 8(1.00794) + 6(15.9994) = 176.12592g/mol

Part B)
500.0mg = 0.5g
Grams to moles, we simply divide the grams by the Molar mass of Vitamin C
0.5 / 176.12592 = 2.8389x10*-3mols of Vitamin C

Part C)
To find the molecules of Vitamin C, we simply multiply the moles by Avogadro's number (6.022x10*23)
2.8389x10*-3 x 6.022x10*23 = 1.7096x10*21 molecules

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