The initial volume in a graduated cylinder was 2.06 mL. A student added 100 drop

Question

The initial volume in a graduated cylinder was 2.06 mL. A student added 100 drops of an oleic acid solution to the graduated cylinder and the final volume was 3.02 mL.

(a) What is the volume of one drop from the dropper? Show full calculation, with units and sig figs.

(b) The solution used was 0.50 % (by volume) oleic acid. What is the volume of oleic acid contained in one drop ? (0.50% = 0.0050) Show full calculation, with units and sig figs.

(c) One of the above drops was used to create a film with a surface area of 116 cm^2 . Assume the film is a very wide but short cylinder, therefore volume of this film = (surface area) x (thickness). Note: The film is created by only the oleic acid in the drop, the remainder of the drop is ethanol, which dissolves into the water. Also remember that 1 mL = 1 cm^3 . (c) Use the volume from (b) and the surface area given to calculate the thickness of the film in cm. Show full calculation, with units and sig figs.

Explanation / Answer

(a) The change in volume of the cylinder is due to the addition of oleic acid; therefore, the volume of 100 drops of oleic acid = (3.02 – 2.06) mL = 0.96 mL.

Therefore, the volume of 1 drop of oleic acid = (0.96 mL/100 drops) = 0.0096 mL/drop (ans).

(b) The solution used is 0.50% oleic acid; therefore, volume of oleic acid in 1 drop = (0.50/100)*(0.0096 mL) = 4.8*10-5 mL (ans).

(c) Let t cm be the thickness of the film.

As per the problem, we have

(surface area)*(thickness of the film) = volume of the film = volume of oleic acid

===> (116 cm2)*(t cm) = 4.8*10-5 mL = (4.8*10-5 mL)*(1 cm3/1 mL) = 4.8*10-5 cm3

===> 116*t cm3 = 4.8*10-5 cm3

===> t = (4.8*10-5)/(116) = 4.1379*10-7 4.14*10-7

The thickness of the film is 4.14*10-7 cm (ans).

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