Question

A sample is prepared by volumetrically transferring 7.85 mL of whiskey into a 50 mL volumetric flask and diluting to the volume mark with the diluent. The whiskey sample is analyzed and the concentration for 3-Methyl-1-butanol (3M1B) is calculated as 13.7 ppm. The concentration of 3M1B in the original whiskey sample is:

2.15 ppm

87.3 ppm

86.6 ppm

2.2 ppm

It's one of these.

Answer 1

Ans. Given, 7.85 mL whiskey is diluted to 50 mL.

Dilution ratio = 7.85 mL / 50 mL = 0.157 : 1

That is, 0.157 mL of whisky is diluted to 1 mL final volume.

Dilution factor = volume of solute / total diluted volume = 0.157

Now,

Total concentration of undiluted sample = (concentration of diluted sample) / dilution factor

                                                = 13.7 ppm / 0.157 = 87.26 ppm

Thus, the concentration of undiluted sample = 87.26 ppm

Correct option: b. 87.3 ppm