Question

The true absorbance for a 1.0 x 10 −5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calculate the percentage by which the apparent concentration deviates from the known concentration.

Answer 1

Answer:

The percentage deviation is  $\Delta M = 1.87$%

Explanation:

From the question we are told that  

     The concentration is of the solution is $C = 1.0*10^{-5} M$

     The true absorbance A = 0.7526

      The percentage of transmittance due to stray light $z = 0.56$% $=\frac{0.56}{100} = 0.0056$

Generally Absorbance is mathematically represented as

           $A = -log T$

Where T is  the percentage of true transmittance

    Substituting value  

          $0.7526 = - log T$

              $T = 10^{-0.7526}$

                  $= 0.177$

                  $= 17.7$%

The Apparent absorbance is mathematically represented

           $A_p = -log (T +z)$

Substituting values

           $A_p = -log(0.177 + 0.0056)$

                $= -log(0.1826)$

               = 0.7385

The percentage by which apparent absorbance deviates from known absorbance is mathematically evaluated as

       $\Delta A = \frac{A -A_p}{A} * \frac{100}{1}$

              $= \frac{0.7526 - 0.7385}{0.7526} * \frac{100}{1}$

             $\Delta A = 1.87$%  

Since Absorbance varies directly with concentration the percentage deviation of the apparent concentration from know concentration  is

              $\Delta M = 1.87$%