Question

6. To make solutions for the Density of KCI graph, you are provided with a 12.0% KCl solution. You pipet 10.00 mL of this solution into a test tube and then you pipet 30.00 mL of deionized water into the same tube. After stirring, what is the KCl concentration of the new solution you've made? This is KCl 4 solution. V snal = 40.00m Csonal=0-03110.00m 80.12=40.000L XC 1: 24am 2 52:0-037P 40.08 18.00 7. Using the equation of the line in the graph below, calculate the concentration of Allura Red (in ppm) of the Buckeye-ade sample which has an absorbance of 0.485 a.u. Absorbance vs. Concentration for a series of Allura Red Solutions 0.900 0.800 0.700 0.600 Absorbance y=0.0471x 12-090 • Absorbance of Allura Red Solutions 0.300 0.200 0.100 0.000 0.00 2.00 4.00 6.00 12.00 14.00 16.00 18.00 0.00 10.00 Concentration (ppm) Page 3 of 3

Answer 1

Beer-Lambert law................ A = &c ...........Eq. 1 where, A = absorbance, a = absorptivity c = concentration, 1 = path lenth if the path lenth, I = 1 cm, then A=&C ...........Eq. 2 we, know the straight line equation y =mx+b...........Eq. 3 where, m = slope, X = concentration b =intercept on y-axis Technically, for Beer-Lambert law, the intercept b should be zero (i.e. zero absorbance at zero concentration). Therefore, the strait line equation-3 becomes y = mx I equivivalent to equation 2. A = EC From the graph, you have the value of slope, m = 0.0471 The absorbance A of unknown compound is given as y = 0.485 a.u. 0.0485 = 0.0471 x x 0.485 = 10.30 ppm * 0.0471 ie, the concentration of unknown sample is 10.30 ppm