Question

Vitamin C was measured by an electrochemical method in a 50.0-mL sample of lemon juice. A detector signal of 2.02 mA was observed. A standard addition of 1.00 mL of 29.4 mM vitamin C increased the signal to 3.79 mA. Find the concentration (in mM units) of vitamin C in the juice.

Answer 1

Answer:

The concentration (in mM units) of vitamin C in the juice is 0.64mM.

Explanation:

Let us calculate -:

Concentration of sample =>$1ml\times 29.4mM$

                               = 29.4 mmol => 29.4mmol/50ml

                               = 0.588mM

plug in formula =>$\frac{[final concentration]}{[v_1/v_2]} \times[final concentration]+[concentration of sample])$ or

$\frac{[x]}{[v_1/v-2]} \times[ x]+[concentration of sample])$

= [2.02 micro-Amber/3.79micro-Amber]

plug in and simplify

1.81x = 1.17

x = 0.64mM

Hence , the answer is 0.64mM.