Calculate and enter the molarity of your three acetic acid trials using the volume of standardized naoh solution required for each and the average molarity of the naoh solution from the standardization trials with khp. you should report 3 significant figures, e. g. 0.488 m.
entry # vol acetic acid(ml) vol na0h(ml) m acetic acid
#1: 5.000 24.80
#2: 5.020 19.20
#3: 5.019 18.00

Answer : The molarity of acetic acid for trial 1, 2 and 3 is 0.374 M, 0.289 M and 0.271 M respectively.

Explanation :

To calculate the molarity of acetic acid the expression used as:

Let us assume that the average molarity of NaOH be 0.0755 M.

For trial 1 :

Volume of NaOH = 24.80 mL

Volume of acetic acid = 5.000 mL

Now put all the given values in the above expression, we get:

For trial 2 :

Volume of NaOH = 19.20 mL

Volume of acetic acid = 5.020 mL

Now put all the given values in the above expression, we get:

For trial 3 :

Volume of NaOH = 18.00 mL

Volume of acetic acid = 5.019 mL

Now put all the given values in the above expression, we get:

he uptake of drugs from the intestine is often strongly influenced by diffusion. for example, consider a water-insoluble steroid for treating inflammation that is solubilized in aggregates of detergent molecules, called micelles. these micelles, aggregates of 80 soap molecules, have a molecular weight of 24,000, a diameter of 2.6 nm, and a charge of -27. the positively charge ions (counterions) in this system are na".