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A vitamin C (ascorbic acid) tablet was dissolved in approximately 50 mL of distilled water and titrated with the standardized NaOH solution. From the results of this titration, the mg of ascorbic acid in the tablet was calculated. Molecular formula of ascorbic acid: C6H8O6 Volume of NaOH required to neutralize ascorbic acid in Vitamin C tablet (mL) 20.74 Concentration of NaOH in mol/L, 0.201 Calculate the amount of ascorbic acid in the Vitamin C tablet in (mg).

Sagot :

Histrionicus

Answer:

The correct answer is - 733.69 mg.

Explanation:

given:

The volume of ascorbic acid = 50 ml

The volume of NaOH = 20.74 ml

Molarity of NaOH = 0.201 M

formula = C6H8O6

Solution:

moles of NaOH for titration = molarity of NaOH × The volume of NaOH

= 0.201 × 20.74 × 10^-3

=4.16874 × 10^-3

then molecular weight of ascorbic acid

= 6 × 12+ 8×1+ 6× 16

=176g/mol

the molecular mass would be

= 4.16874 × 10^-3 × 176

= 733.69 × 10^-3 g

=733.69 mg

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