A student grinds a vitamin C tablet and dissolves it in 250. mL of water in a vo

Question

A student grinds a vitamin C tablet and dissolves it in 250. mL of water in a volumetric flask. The tablet had a mass of 1.075 grams before grinding and 0.992 grams of powder were transferred into the volumetric flask. 30.0 mL of the vitamin C solution is placed into an Erlenmeyer flask and the redox titration is performed. If it takes 9.50 mL of the 0.0151M triodide titrant to reach the equivalence point, how many milligrams of ascorbic acid (MW 176.12) were in the original tablet? Please show work!

Explanation / Answer

Based on data provide by you,

0.992g of vit c in 250mL

so in 30mL its about 0.12g

Based on data provide we need find out moles of solute

molarity = moles of solute / liter of solution

0.0151 = moles of solute / 0.0095

moles of triodate titrant = 0.0095*0.0151 = 0.000143/2 (two iodide reuire per mol of ascorbic acid) = 0.0000715 moles of ascorbic acid

gm of vit c = moles of vit c * mw

gm of vit c = 0.0000715*176.12

gm of vit c = 0.0126g / 30mL

now for 250mL = 0.0126g * 250mL / 30mL = 0.104g in 250mL

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