Question

The density of a 40 wt% solution of ethanol in water is 0.937 g/mL. The density of n-butanol is 0.810 g/mL. What is the concentration of n-butanol in ppm if you dissolve 20 μL of this alcohol in a 40 wt% solution of Ethanol in water. The final volume of the solution is 25 mL. Assume that the density of the ethanol solution does not change upon dissolution of the butanol. Use the correct number of significant figures! Hint: to designate a number like "1000" with 4 significant figures you need to enter a decimal point "1000.".

Answer 1

Ans. Given, density of n-butanol = 0.810 g/ mL                          ; [1 mL = 1000 ug]

Volume of n-butanol taken = 20.0 uL = 0.020 mL

Mass of n-butanol taken = Volume x Density

                                                = (0.020 mL) x (0.810 g/ mL)

                                                = 0.0162 g

# Volume of 40% ethanol taken = Total volume of solution – Volume of n-butanol taken

                                                = 25.0 mL – 0.020.0 mL

                                                = 24.980 mL

Mass of 40% ethanol taken = (24.980 mL) x 0.937 g mL^-1 = 23.40626 g

# Total mass of solution = Mass of n-butanol + Mass of 40% ethanol

                                                = 0.0162 g + 23.40626 g

                                                = 23.42246 g

# Now,

            [n-butanol], ppm = Mass of n-butanol in mg / Mass of solution in kg

                                          = (0.0162 g x 1000 mg g^-1) / (23.42246 g x 10^-3 kg g^-1)

                                          = 16.200 mg/ 0.02342246 kg

                                          = 691.6438 ppm

                                          = 692.0 ppm

Hence, [n-butanol], ppm = 692.0 ppm