(a) [10 pts] Using an electrochemical method, you measured the ascorbic acid of a 30.0 mL watermelon juice sample. A detector signal of 2.12 uA was observed. A standard addition of 1.00 mL of 24.9 mM ascorbic acid standard increased the current to 3.09 uA. Find the concentration of vitamin C in the watermelon juice.

Explanation:

We assume that the concentration of unknown sample is x M.

As the volume is given as 30.0 ml or 0.030 L (as 1 L = 1000 ml) therefore, moles of ascorbic acid will be calculated as follows.

                    =

                    = 0.030x mole

Therefore, standard moles of ascorbic acid added are as follows.

          = mole

As the detector signal is additive in nature so, signal due to mole of ascorbic acid alone will be calculated as follows.

             = 0.97

This shows that for mol, the signa;l observed is 0.97 . Therefore, 1 will be equal to as follows.

       1 =

and, moles that will observe 2.12 signal will be calculated as follows.

           =

Therefore, 0.030 x =

                          x = 1.814 mM

Thus, we can conclude that the concentration of vitamin C in the watermelon juice is 1.814 mM.

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.