Question

the titration of vitamin c with iodine proceeds according to the given equation (c_{6}h_{8}o_{6}+i_{2}
ightarrow c_{6}h_{6}o_{6}+2hi). suppose it takes 23.71 ml of 0.0026 m (i_{2}) solution to reach the end - point of the titration. how many moles of vitamin c are present?

Explanation:

Step1: Calculate moles of I₂

Use the formula $n = M\times V$, where $M$ is molarity and $V$ is volume in liters. First convert volume of $I_2$ solution from mL to L: $V = 23.71\ mL=23.71\times10^{- 3}\ L$, and $M = 0.0026\ M$. Then $n_{I_2}=M\times V=0.0026\ mol/L\times23.71\times10^{-3}\ L$.
$n_{I_2}=0.0026\times23.71\times10^{-3}=6.1646\times10^{-5}\ mol$

Step2: Determine moles of vitamin C

From the balanced chemical equation $C_{6}H_{8}O_{6}+I_{2}
ightarrow C_{6}H_{6}O_{6}+2HI$, the mole - ratio of vitamin C ($C_{6}H_{8}O_{6}$) to $I_2$ is 1:1. So, $n_{C_{6}H_{8}O_{6}}=n_{I_2}$.
$n_{C_{6}H_{8}O_{6}} = 6.1646\times10^{-5}\ mol$

Answer:

$6.16\times10^{-5}\ mol$