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 15 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!

Answer 1

Given

Density 40% of ethanol in water = 0.937 g/mL

Density of n-butanol = 0.810 g/mL

mass of Butanol =74 g/mol

butanol = 15 µL (I.e. in mL = 15 / 1000 =0.015mL )

amount of butanol in g = volume x density

butanol in g = 0.015 x 0.810=0.01215 g of butanol

so now find concentration in ppm n-butanol for 0.01215 g of butanol in 25 mL of solution 40 % ethanol in water

convert 25 mL of solution 40 % ethanol in water in gram

solution 40 % ethanol in water = 25 x 0.937 =23.425 g

solution 40 % ethanol in water = 23.425 g

Now butanol (solute ) =0.01215 g (i.e. 0.01215 x 1000 = 12.15 mg ) & solution = 23.425 g (i.e. in kg = 23.425 / 1000 =0.023425 kg)

Formula for ppm

1 g solute per 1,000,000 g solution or ppm = 1 mg solute per 1 kg solution

ppm = (0.01215 / 23.425 ) x 1000000 =518.68

concentration of Butanol =518.68 ppm in 25 mL of solution ethanol in water