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

Ans :-

Let initial concentration of glucose solution = M₁ and

The concentration after dilution is = M₂

Given , Initial Volume of the solution = V₁ = 3.34 mL

Volume after dilution = V₂ = 100.0 mL

On applying equation of dilution :

M₁V₁ (Before dilution)= M₂V₂ (After dilution)

M₁ (3.34 mL) = M₂ (100.0 mL)

So,

M₂ = M₁ x 3.34 mL / 100.0 mL ..................(1)

Again,

M₁V₁ = M₂V₂

Put value of equation (1) in this equation :

(0.070 M).(100.0 mL) (After dilution dilution)= (M₁ x 3.34 mL / 100.0 mL).(10.0 mL)  (Before dilution)

7 = M₁ (0.334 )

M₁ = 20.96 M

Hence, concentration of glucose in unknown solution = 20.96 M