Question

A student dissolves 0.2062 g of caffeine (MW: 194.19 g/mol) into 50.00 mL of deionized water. What is the concentration of the caffeine in the solution in ppm (parts per million)?

A student performs a spectrophotometric analysis using external calibration and obtains a linear relationship with a slope of 0.0550 and a y-intercept of 0.011. In order to analyze an unknown solution, the student dilutes 10.00 mL of the unknown solution to 50.00 mL. If the absorbance of the dilute unknown solution is measured as 0.350, then what is the concentration of the unknown solution?

Answer 1

1. Mass of caffeine dissolved = 0.2062 g.

Volume of deionized water required to dissolve the caffeine = 50.00 mL.

The concentration of a substance is defined as 1 ppm when 1 mg of the substance is dissolved in 1 L of the solvent.

Hence, concentration of the caffeine solution in ppm =

(0.0262 g)/(50.00 mL)

= [(0.2062 g)*(1000 mg)/(1 g)]/[(50.00 mL)*(1 L)/(1000 mL)]

= (206.2 mg)/(0.05 L)

= 4124 mg/L = 4124 ppm (ans, correct to 4 sig. figs).

2. The calibration curve follows the equation

y = 0.0550x + 0.011

where y denotes the absorbance and x is the concentration of the substance giving absorbance y.

The diluted solution of the unknown gave an absorbance of 0.350.

Plug in values and get

0.350 = 0.0550x + 0.011

=====> 0.0550x = 0.339

=====> x = (0.339)/(0.0550) = 6.1636

The concentration of the diluted solution of the unknown is 6.1636 M (no unit is given; we can assume molarity).

10.00 mL of the unknown solution was diluted to 50.00 mL; hence, dilution factor (DF)

= (50.00 mL)/(10.00 mL)

= 5.00

Concentration of the unknown in the original solution = (concentration in diluted solution)*(DF)

= (6.1636 M)*(5.00)

= 30.818 M

≈ 30.82 M (ans, correct to 4 sig. figs).