Answer:
0.02244 M is the molarity of the iodine solution.
Explanation:
Ascorbic acid +[tex]I_2[/tex] → dehydroascorbic acid + [tex]2H^++2I^_[/tex]
Moles of ascorbic acid = [tex]\frac{0.1000 g}{176 g/mol }=0.0005682 mol[/tex]
According to reaction, 1 mole of ascorbic acid reacts with 1 mole [tex]I_2[/tex] Then 0.0005682 mole of ascorbic caid will :
[tex]\frac{1}{1}\times 0.0005682 mol=0.0005682 mol[/tex] of [tex]I_2[/tex] solution
Moles of [tex]I_2[/tex] solution = 0.0005682 mol
Volume [tex]I_2[/tex] solution = 25.32 mL = 0.02532 L( 1 mL= 0.001 L)
Molarity of the [tex]I_2[/tex] solution :
[tex][I_2]=\frac{0.0005682 mol}{0.02532 L}=0.02244 M[/tex]
0.02244 M is the molarity of the iodine solution.
