Question

An industrial hygenist creates a standard atmosphere by injecting 250 ul of chlorine into a sealed container that is 1ft x 2 ft x 2 ft. The hygenist then measure the air by three different analytical methods; colormetric tube, sobent tube and direct reading instrument. If the results for each of the instrument was a follows: colormetric tube 2.6 ppm, sobent tube 2.0 ppm, direct reading instrument 1.9 ppm, what is the percent error for each of the methods?

Answer 1

Volume of chlorine = 250 ul = 250 x 10^-6 litres        ( 1 ul = 1 microlitre = 1x 10^-6 litre)

Volume of sealed container = 1 ft x 2 ft x 2 ft

                                                     = 4 ft³       

                                              = 4 x 28.3168 litre      ( 1 ft³ = 28.3168 litre)

                                              = 113.2672 litre

Hence 250 x 10^-6 litres        chlorine is present in 113.2672 litre volume of container.

Hence, 113.2672 part volume contains 250 x 10^-6 part chlorine.

So, 1 million (10⁶) part volume contains = (250 x 10^-6) x 10⁶ / 113.2672

                                                      = 2.207 part

Therefore concentration of chlorine in the container ( standard ) = 2.207 ppm

% error = [│(experimental - standard│ / standard] x 100

(i)

colormetric tube     -     exprimental concentration of chlorine = 2.6 ppm

% error = [│(2.6 – 2.207│ / 2.207] x 100

             = 17.80 %

(ii) sobent tube        exprimental concentration of chlorine   =   2.0 ppm

% error = [│(2.0 – 2.207│ / 2.207] x 100

               = 9.38%

(iii) Direct reading instrument     exprimental concentration of chlorine   =   1.9 ppm

% error = [│(1.9 – 2.207│ / 2.207] x 100

            = 13.91 %