Question

1. swirl the flask in order to stir the solution in the erlenmeyer flask for approximately 2 to 5 minutes until all of the acid - powder has dissolved.
2. use a funnel to fill a clean buret to the 25 - ml mark with 0.100 m naoh (sodium hydroxide).
3. place the buret in a buret clamp on the buret stand.
4. record the concentration of the naoh (0.100 m naoh) in the data table.
5. record the initial level of naoh that is in the buret in the data table.
6. place the erlenmeyer flask in the space below the buret so that the naoh can be added.
7. begin the titration by slowly opening the stopcock on the buret so that the naoh is added at a rate of one drop per second. as you titrate, make sure to swirl the flask to stir the contents.
8. continue the titration until the solutions turns a permanent faint pink color and then stop the titration by closing the stopcock on the buret.
9. record the final level of naoh that is in the buret in the data table.
10. if instructed to do a second trial, repeat steps 1 through 13 and record the data in the appropriate places for trial 2 in the data table.

data table

	trial 1	trial 2
concentration of base (naoh)	0.100 m
initial level of naoh	25 ml
final level of naoh	45.3 ml
volume of base used	20.3 ml
molar mass of unknown acid
average molar mass of unknown acid

data analysis

1. we will first find the number of moles of base (naoh) used to neutralize the unknown acid by using the molarity equation: molarity = $\frac{moles of base}{volume of base used (l)}$

Explanation:

Step1: Convert volume to liters

The volume of base used is $20.0\ ml$. Convert it to liters: $V = 20.0\ ml\times\frac{1\ L}{1000\ ml}=0.0200\ L$.

Step2: Calculate moles of base

The molarity of $NaOH$ is $M = 0.100\ M$. Using the molarity formula $M=\frac{n}{V}$ (where $n$ is moles and $V$ is volume), we can find the moles of $NaOH$: $n = M\times V=0.100\ mol/L\times0.0200\ L = 0.00200\ mol$.

Step3: Calculate molar mass of unknown acid

Assume the acid is monoprotic, and the mole - ratio of acid to base at neutralization is $1:1$. The mass of the unknown acid is $m = 0.15\ g$. Using the formula for molar mass $M_{acid}=\frac{m}{n}$, we get $M_{acid}=\frac{0.15\ g}{0.00200\ mol}=75.0\ g/mol$.

Answer:

The molar mass of the unknown acid for Trial 1 is $75.0\ g/mol$