Part A: Using your graduated cylinder, you calibrate the pipet you will be using
ID: 148827 • Letter: P
Question
Part A: Using your graduated cylinder, you calibrate the pipet you will be using and the values you collect are as follow. What is the average number of drops per mL of benzene.
20 drops 19 drops 17 drops
Anwser:
18.7
Part B: After calibration, you use the oleic acid/benzene solution to form a monolayer. It takes 8 drops , 9 drops and 7 drops before the layer on the water solution starts to make a bubble tha does not go into solution. What is the average number of drops for your monolayer?
Part C: Since we have the number of drops it takes to make a monolayer and we have the number of drops it takes to make one mL of benzene, we can determine the mL of oleic acid/benzene it takes to form a monolayer.
Part D: Next we need to determine the mass of oleic acid in the monolayer. The concentration of the oleic acid/benzene solution is 0.02g/L. Calculate how many grams of oleic acid is in one drop.
Part E: With this information, we now need to calculate the volume in mL the drop of oleic acid occupy using the density of 0.895 g/mL and the number of drops of oleic acid solution used.
Part F: Determine the volume of the monolayer from the above information.
18.7
dropsExplanation / Answer
Part A
We have 3 reading for the no. of drops in 1 ml of benzene
20,19,17 drops
Average no. of drops per ml of benzene = sum of all drops/3
= 20+19+17/3
=56/3
=18.7
Part b
We have reading for no. of drops used for forming monolayer
8 drops , 9 drops and 7 drops
Average no. of drops in monolayer = sum of all drops /3
=8+9+7/3
=24/3
=8
Part c
=ml of oleic acid / benzene required to form monolayer
=(1ml /no. of drops in 1ml) x no. of drops in monolayer
=(1/18.7) x 8
= 0.42ml
Part d
Given =0.02g/l
Mass of oleic acid / benzene in monolayer
= given concentration x volume of monolayer
=0.02 x 0.42 x10^-3
=8.4 x 10^-6 g
Part e
Mass of one drop of oleic acid = 8.4 x 10^-6 g / 8
= 1.05 x 10^-6 g
Density = 0.895g/ml
Volume of drop = mass/density
= 1.05 x 10^-6 g/0.895g/ml
=1.17 x 10^-6 ml
No. of drops = volume of monolayer / volume of one drop
=0.42ml/1.17 x 10^-6 ml
=3.6 x10^5
Part f
Density of monolayer =0.895g/ml
Mass of monolayer =mass of 1 drop x no. of drop
= 1.05 x 10^-6 g x 3.6 x10^5
=0.378g
So volume of monolayer
= mass of monolayer /density
=0.378/0.895
=0.4223ml