A 0.500 g vitamin C tablet is dissolved and titrated directly with 0.0500 M iodine solution (starch indicator); 41.0 mL are needed. The reaction is C6H8O6 + I2 -> C6H6O6 + 2HI (ascorbic acid: 176 g/mol). Find the percentage of ascorbic acid in the tablet (one decimal place).

JEE Main levelNumerical answerDifficulty 2 of 3eq141

A 0.500 g vitamin C tablet is dissolved and titrated directly with 0.0500 M iodine solution (starch indicator); 41.0 mL are needed. The reaction is $C X_{6} H X_{8} O X_{6} + I X_{2} C X_{6} H X_{6} O X_{6} + 2 HI$ (ascorbic acid: 176 g/mol). Find the percentage of ascorbic acid in the tablet (one decimal place).

Numerical answer type, answer in %, tolerance 0.06.

I X_{2}

= 0.0500

\times

41.0 = 2.05 mmol, and the stoichiometry is 1:1, so ascorbic acid = 2.05 mmol = 2.05

\times

176 = 360.8 mg. Percentage = 0.3608/0.500

\times

100 = 72.2 %.