Question

Question 1 2 pts 3.34 mL of an unknown glucose solution was diluted in a 100.00 mL of volumetric flask with distilled water. 10mL of that solution and 4.00 mL of the ferricyanide reagent was placed in a second 100.00 mL volumetric flask and diluted to the mark with distilled water. Spectrophotometric analysis of the glucose/ferricyanide solution showed that the concentration of glucose was 0.070 M. What is the concentration of glucose in the unknown solution?

Answer 1

3.34 mL of an unknown solution of glucose was diluted in a 100.00 mL of volumetric flask with distilled water. Thus using the dilution equation,

$M_{1}\plus \times V_{1}=M_{2}\plus \times V_{2}$

Here, $M_{1}$ and $V_{1}$ are the concentration and volume before dilution and $M_{2}$ and $V_{2}$ after dilution.

$3.34 mL\plus \times x=100 mL\plus \times y.............(1)$

Here x is the concentration of the glucose in the unknown solution and y is the concentration in 100 mL volumetric flask.

10 mL of this solution is again diluted in a 100 mL flask.

Thus, $10 mL\plus \times y=100 mL\plus \times z.............(2)$

Here z is the concentration of the glucose solution after second dilution.

Spectrophotometric analysis showed that the concentration of glucose after second dilution is $0.070 M$.

That is, $z=0.070M$

Thus from equation 2,

$10 mL\plus \times y=100 mL\plus \times 0.070M$

$y=\frac{100mL\times 0.070M}{10mL}=0.70M$

Substituting in equation 1,

$3.34 mL\plus \times x=100 mL\plus \times 0.70M$

$x=\frac{100mL\times 0.70M}{3.34mL}=20.9 M$

Thus the concentration of glucose in the unknown solution is 20.9 M.